\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
CommeUnJeu · L1 MPSI

Standard functions

⌚ ~99 min ▢ 12 blocks ✓ 46 exercises ➣ Prerequisites : Differentiability, Real functions
The lycée gives names to a handful of functions that the rest of analysis hangs on: \(\ln\), \(\exp\), \(x \mapsto x^\alpha\), \(\sh\), \(\ch\), \(\th\), and the inverse trigonometric functions \(\arccos\), \(\arcsin\), \(\arctan\). This chapter promotes all of them to fully rigorous objects: explicit domains, regularity (\(C^\infty\) on the appropriate interval), derivative formulas, monotonicity, named identities, asymptotic comparisons. The inverse-function derivative formula --- proved (admitted) in Real functions --- is the workhorse: the regularity and derivative of \(\ln\), \(\arccos\), \(\arcsin\), \(\arctan\) all flow from it. A short final section deals with complex-valued functions of a real variable and the formula \((\mathrm{e}^g)' = g' \, \mathrm{e}^g\) for \(g : I \to \mathbb{C}\) derivable.
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Convention. Algebraic identities are stated on the natural domain of each function (\(\mathbb{R}_+^*\) for \(\ln\), \(\mathbb{R}\) for \(\exp\), \(\sh\), \(\ch\), \(\th\), \([-1, 1]\) for \(\arccos\) and \(\arcsin\), \(\mathbb{R}\) for \(\arctan\)). Derivability and class-\(C^k\) statements live on an interval \(I \subset \mathbb{R}\) with non-empty interior. Throughout, \(\mathrm{e} = \exp(1)\) is the Néper constant.
I Logarithm and exponential
The natural logarithm \(\ln\) is the unique function on \(\mathbb{R}_+^*\) with derivative \(1/x\) vanishing at \(1\). Its inverse \(\exp\) is then defined as the inverse function. The pair turns multiplicative structure into additive structure: \(\ln(xy) = \ln x + \ln y\) and \(\exp(x + y) = \exp(x) \exp(y)\). These functional equations propagate everywhere in analysis, from limits to differential equations.
Definition — Natural logarithm
There exists a unique function \(\textcolor{colordef}{\ln : \mathbb{R}_+^* \to \mathbb{R}}\) of class \(C^\infty\) satisfying $$ \forall x > 0, \quad \ln'(x) = \frac{1}{x} \qquad \text{and} \qquad \ln(1) = 0. $$ This function is called the natural logarithm. (Existence is admitted: it relies on the fundamental theorem of calculus, established in Integration on a segment. Uniqueness is immediate --- the difference of two such functions has zero derivative on the interval \(\mathbb{R}_+^*\), hence is constant, and the constant is \(0\) by evaluation at \(1\).)
Proposition — Functional equation of \(\ln\)
For every \(x, y > 0\) and every \(n \in \mathbb{Z}\): $$ \textcolor{colorprop}{\ln(xy) = \ln x + \ln y, \quad \ln\!\left(\frac{1}{x}\right) = -\ln x, \quad \ln\!\left(\frac{x}{y}\right) = \ln x - \ln y, \quad \ln(x^n) = n \ln x.} $$

Fix \(y > 0\) and consider the function \(\varphi_y : \mathbb{R}_+^* \to \mathbb{R}\) defined by \(\varphi_y(x) = \ln(xy) - \ln(x)\). By the chain rule: $$ \varphi_y'(x) = \frac{1}{xy} \cdot y - \frac{1}{x} = \frac{1}{x} - \frac{1}{x} = 0. $$ So \(\varphi_y\) is constant on the interval \(\mathbb{R}_+^*\). Evaluating at \(x = 1\): \(\varphi_y(1) = \ln(y) - \ln(1) = \ln y\). Hence \(\ln(xy) - \ln x = \ln y\) for every \(x > 0\), that is, \(\ln(xy) = \ln x + \ln y\). The remaining identities follow:
  • \(\ln(x \cdot 1/x) = \ln 1 = 0\), so \(\ln(1/x) = -\ln x\).
  • \(\ln(x/y) = \ln(x \cdot 1/y) = \ln x + \ln(1/y) = \ln x - \ln y\).
  • For \(n \ge 0\), by induction: \(\ln(x^{n+1}) = \ln(x^n \cdot x) = \ln(x^n) + \ln x = n \ln x + \ln x = (n+1) \ln x\). For \(n < 0\), write \(\ln(x^n) = \ln(1/x^{-n}) = -\ln(x^{-n}) = -(-n)\ln x = n \ln x\).

Proposition — Variations and limits of \(\ln\)
The function \(\ln\) is strictly increasing on \(\mathbb{R}_+^*\), and $$ \textcolor{colorprop}{\lim_{x \to 0^+} \ln x = -\infty, \qquad \lim_{x \to +\infty} \ln x = +\infty.} $$ In particular, \(\ln\) realizes a bijection from \(\mathbb{R}_+^*\) onto \(\mathbb{R}\).

  • Strict monotonicity. Since \(\ln'(x) = 1/x > 0\) on \(\mathbb{R}_+^*\), \(\ln\) is strictly increasing.
  • Limit at \(+\infty\). Since \(2 > 1\) and \(\ln\) is strictly increasing, \(\ln 2 > \ln 1 = 0\). By P1.1, \(\ln(2^n) = n \ln 2\) for every \(n \in \mathbb{N}\), hence \(\ln(2^n) \to +\infty\). By monotonicity, for every \(x \ge 2^n\), \(\ln x \ge n \ln 2\), hence \(\ln x \to +\infty\) as \(x \to +\infty\).
  • Limit at \(0^+\). As \(x \to 0^+\), \(1/x \to +\infty\), and \(\ln x = -\ln(1/x) \to -\infty\) (P1.1).
  • Bijectivity. A strictly increasing continuous function on the interval \(\mathbb{R}_+^*\) with limits \(-\infty\) and \(+\infty\) at the endpoints is a bijection onto \(\mathbb{R}\) (intermediate-value-theorem corollary from Real functions).

Definition — Exponential\(\virgule\) Néper constant
Since \(\ln : \mathbb{R}_+^* \to \mathbb{R}\) is a bijection (P1.2), it admits an inverse, called the exponential function and denoted \(\textcolor{colordef}{\exp : \mathbb{R} \to \mathbb{R}_+^*}\). It is characterized by $$ \forall x \in \mathbb{R}, \forall y > 0, \quad y = \exp(x) \iff x = \ln(y). $$ The Néper constant is \(\textcolor{colordef}{\mathrm{e} = \exp(1)}\), the unique real number with \(\ln \mathrm{e} = 1\). The exponential is of class \(C^\infty\) on \(\mathbb{R}\) with \(\exp' = \exp\) and \(\exp(0) = 1\) (admitted; consequence of the inverse-function differentiation theorem from Real functions, since \(\ln' = 1/x\) does not vanish on \(\mathbb{R}_+^*\)).
Proposition — Functional equation of \(\exp\)
For every \(x, y \in \mathbb{R}\) and every \(n \in \mathbb{Z}\): $$ \textcolor{colorprop}{\exp(x + y) = \exp(x) \exp(y), \quad \exp(-x) = \frac{1}{\exp(x)}, \quad \exp(nx) = \exp(x)^n.} $$

Apply \(\ln\) to both sides of \(\exp(x + y) = \exp(x) \exp(y)\) and check that they have the same image:
  • \(\ln(\exp(x + y)) = x + y\) (since \(\ln \circ \exp = \mathrm{Id}_\mathbb{R}\)).
  • \(\ln(\exp(x) \exp(y)) = \ln(\exp(x)) + \ln(\exp(y)) = x + y\) (P1.1 functional equation of \(\ln\), then \(\ln \circ \exp = \mathrm{Id}\)).
So both sides have the same image under \(\ln\). By injectivity of \(\ln\) (P1.2 says \(\ln\) is strictly increasing, hence injective), the two sides are equal: \(\exp(x + y) = \exp(x) \exp(y)\). The remaining identities follow from \(\exp(0) = 1\) and induction on \(n\), exactly as in the proof of P1.1.

Proposition — Variations and limits of \(\exp\)
The function \(\exp\) is strictly increasing on \(\mathbb{R}\), and $$ \textcolor{colorprop}{\lim_{x \to -\infty} \exp(x) = 0, \qquad \lim_{x \to +\infty} \exp(x) = +\infty.} $$ Moreover \(\exp(\mathbb{R}) = \mathbb{R}_+^*\). (The result is a direct consequence of \(\exp\) being the inverse of the increasing bijection \(\ln : \mathbb{R}_+^* \to \mathbb{R}\) from P1.2.)
Example
Compute \(\ln\!\bigl( \mathrm{e}^3 \cdot \mathrm{e}^{-2} / \mathrm{e} \bigr)\).

By P1.1: \(\ln(\mathrm{e}^3 \cdot \mathrm{e}^{-2} / \mathrm{e}) = \ln(\mathrm{e}^3) + \ln(\mathrm{e}^{-2}) - \ln(\mathrm{e}) = 3 - 2 - 1 = 0\). (Equivalently: the argument equals \(\mathrm{e}^{3-2-1} = \mathrm{e}^0 = 1\), and \(\ln 1 = 0\).)

Example
Solve on \(\mathbb{R}_+^*\): \(\ln(x) + \ln(x + 1) = \ln 6\).

Both sides exist for \(x > 0\) and \(x + 1 > 0\), that is, \(x > 0\). By P1.1, the equation rewrites as \(\ln(x(x+1)) = \ln 6\), hence by injectivity of \(\ln\): $$ x(x+1) = 6 \iff x^2 + x - 6 = 0 \iff (x - 2)(x + 3) = 0. $$ The roots are \(x = 2\) and \(x = -3\); only \(x = 2\) lies in \(\mathbb{R}_+^*\). So the unique solution is \(x = 2\).

Example
Verify \(\ln(\mathrm{e}) = 1\) from the definition of \(\mathrm{e}\), and check that \(\exp(\ln 5) = 5\).

The definition \(\mathrm{e} = \exp(1)\) means: \(\mathrm{e}\) is the unique real with \(\ln \mathrm{e} = 1\). So \(\ln(\mathrm{e}) = 1\) holds by definition. For the second: \(\exp\) is the inverse of \(\ln\), hence \(\exp \circ \ln = \mathrm{Id}_{\mathbb{R}_+^*}\), in particular \(\exp(\ln 5) = 5\). (Conversely \(\ln \circ \exp = \mathrm{Id}_\mathbb{R}\), so \(\ln(\exp x) = x\) for every \(x \in \mathbb{R}\).)

Example — Graphs of \(\ln\) and \(\exp\)
The graphs of \(\ln\) and \(\exp\) are symmetric about the line \(y = x\) (a general fact about a function and its inverse). The graph of \(\exp\) passes through \((0, 1)\) and lies above \(y = x + 1\); the graph of \(\ln\) passes through \((1, 0)\) and lies below \(y = x - 1\) (a consequence of the fundamental inequalities below).
Proposition — Fundamental inequalities
$$ \textcolor{colorprop}{\forall x \in \mathbb{R}, \quad \exp(x) \ge 1 + x,} \qquad \text{with equality iff } x = 0. $$ $$ \textcolor{colorprop}{\forall x > -1, \quad \ln(1 + x) \le x,} \qquad \text{with equality iff } x = 0. $$

  • Step 1 --- exponential inequality. Set \(\varphi(x) = \exp(x) - 1 - x\) on \(\mathbb{R}\). Then \(\varphi'(x) = \exp(x) - 1\), which is negative for \(x < 0\) and positive for \(x > 0\) (since \(\exp\) is strictly increasing and \(\exp(0) = 1\)). So \(\varphi\) is strictly decreasing on \(]-\infty, 0]\) and strictly increasing on \([0, +\infty[\), with global minimum \(\varphi(0) = 1 - 1 - 0 = 0\). Hence \(\varphi(x) \ge 0\) for every \(x \in \mathbb{R}\), with equality iff \(x = 0\). This gives \(\exp(x) \ge 1 + x\).
  • Step 2 --- logarithmic inequality. For \(x > -1\), \(1 + x > 0\), so \(\ln(1+x)\) is defined. Apply Step 1 with \(\ln(1+x)\) in place of \(x\): \(\exp(\ln(1+x)) \ge 1 + \ln(1+x)\), that is, \(1 + x \ge 1 + \ln(1+x)\), hence \(\ln(1+x) \le x\). Equality holds iff \(\ln(1+x) = 0\), that is, \(x = 0\).

Skills to practice
  • Solving log/exp equations and inequalities
  • Proving inequalities via the fundamental log/exp inequalities
II Power functions
For \(x > 0\) and \(\alpha \in \mathbb{R}\), the power \(x^\alpha\) is defined by \(x^\alpha = \exp(\alpha \ln x)\). The construction generalizes integer powers (where \(x^n = x \cdot x \cdots x\), \(n\) factors) to real exponents and is the natural extension once \(\ln\) and \(\exp\) are available. The algebraic properties \(x^{\alpha + \beta} = x^\alpha x^\beta\), \((x^\alpha)^\beta = x^{\alpha\beta}\) are direct consequences of the functional equations of \(\ln\) and \(\exp\).
Definition — Real power
For \(x > 0\) and \(\alpha \in \mathbb{R}\), the real power of \(x\) with exponent \(\alpha\) is $$ \textcolor{colordef}{x^\alpha = \exp(\alpha \ln x).} $$ The function \(x \mapsto x^\alpha\) is defined:
  • on \(\mathbb{R}_+^*\) in general;
  • on \(\mathbb{R}_+ = [0, +\infty[\) if \(\alpha > 0\), by setting \(0^\alpha = 0\) (continuity extension);
  • on \(\mathbb{R}\) if \(\alpha \in \mathbb{N}\), with \(x^n = x \cdot x \cdots x\) (\(n\) factors), \(x^0 = 1\).
For an integer exponent \(\alpha = n \in \mathbb{Z}\), the formula \(x^n = \exp(n \ln x)\) on \(\mathbb{R}_+^*\) matches the usual convention: for \(n \ge 0\), the iterated product \(x \cdot x \cdots x\) (\(n\) factors), and for \(n < 0\), the reciprocal \(x^n = 1/x^{-n}\) (this is the consistency check from P1.1: \(\ln(x^n) = n \ln x\)).
Proposition — Algebraic properties of powers
For \(x, x' > 0\) and \(\alpha, \beta \in \mathbb{R}\): $$ \textcolor{colorprop}{\begin{aligned} & x^{\alpha + \beta} = x^\alpha \, x^\beta, \qquad (x^\alpha)^\beta = x^{\alpha \beta}, \qquad (x x')^\alpha = x^\alpha \, x'^\alpha, \\ & x^{-\alpha} = \frac{1}{x^\alpha}, \qquad \ln(x^\alpha) = \alpha \ln x. \end{aligned}} $$

All five identities follow from the definition \(x^\alpha = \exp(\alpha \ln x)\) combined with the functional equations of \(\ln\) (P1.1) and \(\exp\) (P1.3). For instance:
  • \(x^{\alpha + \beta} = \exp((\alpha + \beta) \ln x) = \exp(\alpha \ln x + \beta \ln x) = \exp(\alpha \ln x) \cdot \exp(\beta \ln x) = x^\alpha \, x^\beta\).
  • \(\ln(x^\alpha) = \ln(\exp(\alpha \ln x)) = \alpha \ln x\) (since \(\ln \circ \exp = \mathrm{Id}_\mathbb{R}\)).
  • \((x^\alpha)^\beta = \exp(\beta \ln(x^\alpha)) = \exp(\beta \cdot \alpha \ln x) = x^{\alpha \beta}\), using the previous identity.
  • \((xx')^\alpha = \exp(\alpha \ln(xx')) = \exp(\alpha (\ln x + \ln x')) = \exp(\alpha \ln x) \exp(\alpha \ln x') = x^\alpha x'^\alpha\).
  • \(x^{-\alpha} = \exp(-\alpha \ln x) = 1 / \exp(\alpha \ln x) = 1/x^\alpha\).

Proposition — Variations and derivative of \(x \mapsto x^\alpha\)
For every \(\alpha \in \mathbb{R}\), the function \(f_\alpha : x \mapsto x^\alpha\) is of class \(C^\infty\) on \(\mathbb{R}_+^*\), with $$ \textcolor{colorprop}{f_\alpha'(x) = \alpha \, x^{\alpha - 1} \quad \text{for every } x > 0.} $$ The sense of variation:
  • if \(\alpha > 0\): \(f_\alpha\) is strictly increasing on \(\mathbb{R}_+^*\);
  • if \(\alpha < 0\): \(f_\alpha\) is strictly decreasing on \(\mathbb{R}_+^*\);
  • if \(\alpha = 0\): \(f_\alpha\) is constant equal to \(1\).
(The convexity profile of \(x \mapsto x^\alpha\) is studied in the chapter Convexity; we do not state it here.)
Example
Compute \(\sqrt[3]{8}\) via the power formula.

\(\sqrt[3]{8} = 8^{1/3} = \exp(\ln 8 / 3) = \exp(\ln(2^3)/3) = \exp(3 \ln 2 / 3) = \exp(\ln 2) = 2\). The cube root of \(8\) is \(2\), in agreement with the lycée definition.

Example
For \(x > 0\) with \(x \ne 1\), simplify \(\bigl(x^{\ln x}\bigr)^{1/\ln x}\).

By P2.1, \((x^\alpha)^\beta = x^{\alpha \beta}\), so \(\bigl(x^{\ln x}\bigr)^{1/\ln x} = x^{\ln x \cdot 1/\ln x} = x^1 = x\) (valid for \(x > 0\) and \(x \ne 1\), so that \(\ln x \ne 0\)).

Example
Compute \(2^\pi\) in the form \(\exp(\cdot)\), and give a numerical estimate using \(\ln 2 \simeq 0{,}693\).

By definition, \(2^\pi = \exp(\pi \ln 2)\). Numerically, \(\pi \ln 2 \simeq 3{,}14159 \times 0{,}693 \simeq 2{,}177\), so \(2^\pi \simeq \exp(2{,}177) \simeq 8{,}82\). (The value is defined through the formula \(2^\pi = \exp(\pi \ln 2)\); no elementary exact form is expected.)

Example — Logarithm in base \(a\)
For \(a > 0\), \(a \ne 1\) and \(x > 0\), the logarithm in base \(a\) is \(\log_a x = \ln x / \ln a\). Show that \(a^{\log_a x} = x\).

By definition: \(a^{\log_a x} = \exp(\log_a x \cdot \ln a) = \exp\!\bigl( (\ln x / \ln a) \cdot \ln a \bigr) = \exp(\ln x) = x\). So \(\log_a\) is the inverse of \(y \mapsto a^y\) on \(\mathbb{R}_+^*\). The case \(a = \mathrm{e}\) recovers \(\log_\mathrm{e} = \ln\); the case \(a = 10\) gives the decimal logarithm \(\log_{10}\) (often written \(\log\)).

Example — Graphs of \(x \mapsto x^\alpha\)
All graphs of \(x \mapsto x^\alpha\) pass through \((1, 1)\). For \(\alpha > 0\) they are increasing; for \(\alpha < 0\) decreasing. The case \(\alpha = 0\) is the horizontal line \(y = 1\).
Method — Differentiating \(x \mapsto u(x)^{v(x)}\)
When the exponent is non-constant, the formula \((x^\alpha)' = \alpha x^{\alpha - 1}\) does not apply directly. Hypotheses. Let \(I \subset \mathbb{R}\) be an interval and \(u, v : I \to \mathbb{R}\) be differentiable, with \(u(x) > 0\) for every \(x \in I\). The right move is to write the function as an exponential: $$ u(x)^{v(x)} = \exp\!\bigl( v(x) \, \ln u(x) \bigr), $$ then differentiate the composite using \((\exp \circ f)' = f' \exp \circ f\): $$ \bigl( u^v \bigr)' = \bigl( v' \, \ln u + v \cdot u'/u \bigr) \cdot u^v. $$ This handles all « variable-exponent » expressions \(x^x\) on \(\mathbb{R}_+^*\), \(x^{\sin x}\) on \(\mathbb{R}_+^*\), \((\ln x)^x\) on \(]1, +\infty[\), etc.
Skills to practice
  • Computing real powers and simplifying expressions
  • Differentiating \(u^v\) functions
III Comparative growth
At \(+\infty\), the three families \(\ln\), \(x^\alpha\) (\(\alpha > 0\)), \(\exp\) grow at completely different rates: the exponential dominates every power, every power dominates every power of the logarithm. In symbols: \(\ln \ll x^\alpha \ll \exp\) at \(+\infty\). The corresponding statements at \(0^+\) govern the rate at which \(\ln\) goes to \(-\infty\) versus negative powers. These growth comparisons are the daily bread of this course: limits, sequences, asymptotic analysis all rest on them.
Theorem — Comparative growth
For every \(\alpha > 0\) and every \(\beta \in \mathbb{R}\): $$ \textcolor{colorprop}{\lim_{x \to +\infty} \frac{(\ln x)^\beta}{x^\alpha} = 0, \qquad \lim_{x \to +\infty} \frac{x^\alpha}{\mathrm{e}^x} = 0, \qquad \lim_{x \to 0^+} x^\alpha \, |\ln x|^\beta = 0.} $$

  • Step 1 --- key inequality \(\ln u \le 2(\sqrt{u} - 1)\) for \(u \ge 1\). Apply the fundamental inequality \(\ln(1 + x) \le x\) for \(x > -1\) to \(x \leftarrow \sqrt{u} - 1\), valid since \(\sqrt{u} - 1 \ge 0 > -1\) for \(u \ge 1\): $$ \ln(\sqrt{u}) = \ln(1 + (\sqrt{u} - 1)) \le \sqrt{u} - 1. $$ Multiplying by \(2\): \(\ln u = 2 \ln \sqrt{u} \le 2(\sqrt{u} - 1) \le 2 \sqrt{u}\).
  • Step 2 --- first limit. For \(u \ge 1\): \(0 \le \ln u / u \le 2 / \sqrt{u} \to 0\) as \(u \to +\infty\). So \(\ln u / u \to 0\). Three cases on \(\beta\):
    • \(\beta = 0\): \((\ln x)^0 / x^\alpha = 1/x^\alpha \to 0\) since \(\alpha > 0\).
    • \(\beta > 0\): with the substitution \(u = x^{\alpha/\beta}\), write $$ \frac{(\ln x)^\beta}{x^\alpha} = \left( \frac{\ln x}{x^{\alpha/\beta}} \right)^\beta = \left( \frac{\beta}{\alpha} \cdot \frac{\ln u}{u} \right)^\beta. $$ As \(x \to +\infty\), \(u \to +\infty\), so \(\ln u / u \to 0\), hence \((\ln x)^\beta / x^\alpha \to 0\).
    • \(\beta < 0\): numerator \((\ln x)^\beta = 1/(\ln x)^{|\beta|} \to 0\) as \(x \to +\infty\), denominator \(x^\alpha \to +\infty\), so the ratio tends to \(0\).
  • Step 3 --- second limit. Set \(u = \mathrm{e}^x\), equivalently \(x = \ln u\). As \(x \to +\infty\), \(u \to +\infty\), and \(x^\alpha / \mathrm{e}^x = (\ln u)^\alpha / u\). By Step 2 (with \(\beta = \alpha\)), this tends to \(0\).
  • Step 4 --- third limit (at \(0^+\)). Substitute \(x = 1/y\). As \(x \to 0^+\), \(y \to +\infty\), and $$ x^\alpha \, |\ln x|^\beta = \frac{|\ln(1/y)|^\beta}{y^\alpha} = \frac{(\ln y)^\beta}{y^\alpha} \to 0 \quad (\text{Step 2}). $$

Example
Compute \(\displaystyle \lim_{x \to +\infty} \frac{x^3}{\mathrm{e}^x}\).

Direct application of the second limit of T3.1 with \(\alpha = 3\): \(x^3 / \mathrm{e}^x \to 0\). (The polynomial growth \(x^3\) is beaten by the exponential.)

Example
Compute \(\displaystyle \lim_{x \to 0^+} x \ln x\).

For \(x \in ]0, 1[\), \(\ln x < 0\), so \(x \ln x = -x \, |\ln x|\). By the third limit of T3.1 with \(\alpha = 1\) and \(\beta = 1\): \(x \, |\ln x| \to 0\) as \(x \to 0^+\), hence \(x \ln x \to 0\).

Example
Compute \(\displaystyle \lim_{x \to +\infty} \frac{x^x}{\mathrm{e}^x}\).

Write \(x^x / \mathrm{e}^x = \exp(x \ln x - x) = \exp\!\bigl( x(\ln x - 1) \bigr)\). As \(x \to +\infty\), \(\ln x - 1 \to +\infty\) and \(x \to +\infty\), so \(x(\ln x - 1) \to +\infty\), hence \(\exp(x(\ln x - 1)) \to +\infty\). Conclusion: \(x^x / \mathrm{e}^x \to +\infty\). (The function \(x^x\) dominates \(\mathrm{e}^x\) at \(+\infty\).)

Example
Compute \(\displaystyle \lim_{x \to +\infty} \frac{(\ln x)^{100}}{x}\).

Direct application of the first limit of T3.1 with \(\alpha = 1\) and \(\beta = 100\): \((\ln x)^{100} / x \to 0\). The exponent \(100\) does not change the conclusion --- any power of the logarithm is beaten by any positive power of \(x\), however small.

Method — Comparing \(f\) and \(g\) at \(+\infty\) via croissances comparées
When both \(f, g \to \pm\infty\) as \(x \to +\infty\), the ratio \(f(x) / g(x)\) has an indeterminate form. To break it, identify the dominant family on each side using the hierarchy $$ \ln \ll x^\alpha \ll \exp \quad (\alpha > 0). $$ Then the ratio simplifies: factor out the dominant block from numerator and denominator, and conclude. Example: \(\dfrac{x^3 + \ln x}{\mathrm{e}^x - x^2}\). The dominant top is \(x^3\) (since \(\ln x \ll x^3\)), the dominant bottom is \(\mathrm{e}^x\) (since \(x^2 \ll \mathrm{e}^x\)). So $$ \frac{x^3 + \ln x}{\mathrm{e}^x - x^2} \sim \frac{x^3}{\mathrm{e}^x} \to 0 \quad \text{(T3.1, second limit)}. $$
Skills to practice
  • Computing limits via croissances comparées
  • Determining asymptotic dominance
IV Hyperbolic functions
The hyperbolic cosine \(\ch\) and sine \(\sh\) are the even and odd parts of the exponential: \(\ch(x) = (\mathrm{e}^x + \mathrm{e}^{-x})/2\), \(\sh(x) = (\mathrm{e}^x - \mathrm{e}^{-x})/2\). They satisfy the fundamental identity \(\ch^2(x) - \sh^2(x) = 1\), the hyperbolic counterpart of \(\cos^2 + \sin^2 = 1\), and the map \(t \mapsto (\ch t, \sh t)\) parametrizes the right branch of the hyperbola \(X^2 - Y^2 = 1\) (since \(\ch t \ge 1\)), as the trigonometric functions parametrize the circle. The third hyperbolic function, \(\th = \sh / \ch\), is defined and bounded on \(\mathbb{R}\).
Definition — Hyperbolic cosine\(\virgule\) sine and tangent
For every \(x \in \mathbb{R}\): $$ \textcolor{colordef}{\ch(x) = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}, \qquad \sh(x) = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}, \qquad \th(x) = \frac{\sh(x)}{\ch(x)}.} $$ The function \(\ch\) is the hyperbolic cosine, \(\sh\) the hyperbolic sine, \(\th\) the hyperbolic tangent. Note that \(\ch(x) \ge 1 > 0\) for every \(x\), so \(\th\) is defined on \(\mathbb{R}\) (no division-by-zero issue, unlike the trigonometric tangent).
Proposition — Parity\(\virgule\) regularity\(\virgule\) derivatives\(\virgule\) fundamental identity
The functions \(\ch\), \(\sh\), \(\th\) are of class \(C^\infty\) on \(\mathbb{R}\), with $$ \textcolor{colorprop}{\ch'(x) = \sh(x), \qquad \sh'(x) = \ch(x), \qquad \th'(x) = 1 - \th^2(x) = \frac{1}{\ch^2(x)}.} $$ Parity: \(\ch\) is even, \(\sh\) and \(\th\) are odd. Variations: \(\sh\) and \(\th\) are strictly increasing on \(\mathbb{R}\); \(\ch\) is strictly decreasing on \(]-\infty, 0]\) and strictly increasing on \([0, +\infty[\). Image: \(\ch(\mathbb{R}) = [1, +\infty[\), \(\sh(\mathbb{R}) = \mathbb{R}\), \(\th(\mathbb{R}) = ]-1, 1[\). Fundamental identity: $$ \textcolor{colorprop}{\forall x \in \mathbb{R}, \quad \ch^2(x) - \sh^2(x) = 1.} $$

  • Regularity and derivatives. \(\ch\) and \(\sh\) are linear combinations of \(x \mapsto \mathrm{e}^x\) and \(x \mapsto \mathrm{e}^{-x}\), both of class \(C^\infty\) on \(\mathbb{R}\). Direct computation: $$ \ch'(x) = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = \sh(x), \qquad \sh'(x) = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = \ch(x). $$ For \(\th = \sh / \ch\), the quotient rule gives $$ \th'(x) = \frac{\ch^2(x) - \sh^2(x)}{\ch^2(x)} = \frac{1}{\ch^2(x)} $$ using the fundamental identity (next bullet). The other form \(\th' = 1 - \th^2\) follows by writing \(\ch^2 - \sh^2 = \ch^2 (1 - \sh^2/\ch^2) = \ch^2(1 - \th^2)\), hence \(1/\ch^2 = (1 - \th^2)\) once divided by \(\ch^2\).
  • Fundamental identity. A direct computation: $$ \ch^2(x) - \sh^2(x) = (\ch(x) + \sh(x))(\ch(x) - \sh(x)) = \mathrm{e}^x \cdot \mathrm{e}^{-x} = 1. $$ (We used \(\ch + \sh = \mathrm{e}^x\) and \(\ch - \sh = \mathrm{e}^{-x}\).)
  • Parity. \(\ch(-x) = (\mathrm{e}^{-x} + \mathrm{e}^x)/2 = \ch(x)\) (even); \(\sh(-x) = (\mathrm{e}^{-x} - \mathrm{e}^x)/2 = -\sh(x)\) (odd); \(\th(-x) = \sh(-x)/\ch(-x) = -\sh(x)/\ch(x) = -\th(x)\) (odd).
  • Variations. \(\ch'(x) = \sh(x)\); the sign of \(\sh\) matches the sign of \(\mathrm{e}^x - \mathrm{e}^{-x}\), which is negative for \(x < 0\) and positive for \(x > 0\). Hence \(\ch\) decreases on \(]-\infty, 0]\), increases on \([0, +\infty[\). \(\sh'(x) = \ch(x) \ge 1 > 0\), so \(\sh\) is strictly increasing. \(\th'(x) = 1/\ch^2(x) > 0\), so \(\th\) is strictly increasing.
  • Limits and image. As \(x \to +\infty\), \(\mathrm{e}^x \to +\infty\), \(\mathrm{e}^{-x} \to 0\), so \(\ch(x), \sh(x) \to +\infty\) and \(\th(x) = (\mathrm{e}^x - \mathrm{e}^{-x})/(\mathrm{e}^x + \mathrm{e}^{-x}) \to 1\). By parity, \(\ch(x) \to +\infty\), \(\sh(x) \to -\infty\), \(\th(x) \to -1\) at \(-\infty\). Continuity then gives the images: \(\ch(\mathbb{R}) = [1, +\infty[\), \(\sh(\mathbb{R}) = \mathbb{R}\), \(\th(\mathbb{R}) = ]-1, 1[\).

Example
Compute \(\ch(\ln 2)\) and \(\sh(\ln 2)\), and verify the fundamental identity.

Since \(\mathrm{e}^{\ln 2} = 2\) and \(\mathrm{e}^{-\ln 2} = 1/2\): $$ \ch(\ln 2) = \frac{2 + 1/2}{2} = \frac{5}{4}, \qquad \sh(\ln 2) = \frac{2 - 1/2}{2} = \frac{3}{4}. $$ Verification: \(\ch^2(\ln 2) - \sh^2(\ln 2) = 25/16 - 9/16 = 16/16 = 1\) \(\checkmark\).

Example
Solve on \(\mathbb{R}\): \(\ch(x) = 2\).

Set \(u = \mathrm{e}^x > 0\). Then \(\ch(x) = (u + 1/u)/2 = 2\), that is, \(u + 1/u = 4\), hence \(u^2 - 4u + 1 = 0\). The discriminant is \(16 - 4 = 12\), so $$ u = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}. $$ Both roots are positive (since \(\sqrt{3} < 2\)), so both are admissible. Hence \(\mathrm{e}^x \in \{2 + \sqrt{3}, 2 - \sqrt{3}\}\), and \(x \in \{\ln(2 + \sqrt{3}), \ln(2 - \sqrt{3})\}\). By symmetry, \(\ln(2 - \sqrt{3}) = -\ln(2 + \sqrt{3})\) (consistent with \(\ch\) being even). The two solutions are \(x = \pm \ln(2 + \sqrt{3})\).

Example — Graphs of \(\ch\)\(\virgule\) \(\sh\)\(\virgule\) \(\th\)
\(\ch\) is symmetric about the \(y\)-axis with minimum \(\ch(0) = 1\). \(\sh\) passes through the origin with \(\sh'(0) = 1\). \(\th\) is bounded between the asymptotes \(y = \pm 1\), with \(\th(0) = 0\).
Out-of-program reminder
The inverse hyperbolic functions are hors programme at this level; the only required identity is \(\ch^2(x) - \sh^2(x) = 1\). So the inverse hyperbolic functions \(\mathrm{argsh}\), \(\mathrm{argch}\), \(\mathrm{argth}\) are not introduced in this chapter; only the fundamental identity is required. (For the curious: \(\ch\) restricted to \([0, +\infty[\) admits an inverse, but this is second-year material.)
Method — Linearizing \(\ch^p \ \sh^q\)
A product of powers of \(\ch\) and \(\sh\) can be written as a linear combination of \(\ch(kx)\) and \(\sh(kx)\) for various integers \(k\), by going back to the \(\mathrm{e}^x, \mathrm{e}^{-x}\) form: $$ \ch(x) = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}, \qquad \sh(x) = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}. $$ Expand the product, collect \(\mathrm{e}^{kx}\) terms, then re-pair them as \(\ch\)/\(\sh\). The technique is the hyperbolic counterpart of the trigonometric linearization \((\cos x)(\sin x) = \frac{1}{2}\sin(2x)\). Example. \(\ch^2(x) = \bigl( (\mathrm{e}^x + \mathrm{e}^{-x})/2 \bigr)^2 = (\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x})/4 = \frac{1}{2}\ch(2x) + \frac{1}{2}\).
Skills to practice
  • Linearizing hyperbolic expressions
  • Solving equations \(\ch x \equal c\)\(\virgule\) \(\sh x \equal c\)
V Inverse trigonometric functions
The functions \(\cos\), \(\sin\), \(\tan\) are not bijective on \(\mathbb{R}\) (they are periodic). To define a reciprocal one must restrict the domain to a maximal interval of monotonicity. The standard conventions are: \(\cos\) on \([0, \pi]\), \(\sin\) on \([-\pi/2, \pi/2]\), \(\tan\) on \(]-\pi/2, \pi/2[\). The inverse functions are \(\arccos\), \(\arcsin\), \(\arctan\). Their derivatives flow from the inverse-function differentiation theorem of Real functions, applied on the open interior where the derivative of the original function is non-zero.
Definition — \(\arccos\)
The restriction \(\cos|_{[0, \pi]} : [0, \pi] \to [-1, 1]\) is continuous and strictly decreasing (admitted; established in Trigonometry), hence bijective. Its inverse is the arccosine: $$ \textcolor{colordef}{\arccos : [-1, 1] \to [0, \pi].} $$ By definition, for \(x \in [-1, 1]\), \(\arccos(x)\) is the unique \(\theta \in [0, \pi]\) such that \(\cos \theta = x\).
Definition — \(\arcsin\)
The restriction \(\sin|_{[-\pi/2, \pi/2]} : [-\pi/2, \pi/2] \to [-1, 1]\) is continuous and strictly increasing, hence bijective. Its inverse is the arcsine: $$ \textcolor{colordef}{\arcsin : [-1, 1] \to [-\pi/2, \pi/2].} $$ For \(x \in [-1, 1]\), \(\arcsin(x)\) is the unique \(\theta \in [-\pi/2, \pi/2]\) such that \(\sin \theta = x\).
Definition — \(\arctan\)
The restriction \(\tan|_{]-\pi/2, \pi/2[} : ]-\pi/2, \pi/2[ \to \mathbb{R}\) is continuous and strictly increasing, hence bijective. Its inverse is the arctangent: $$ \textcolor{colordef}{\arctan : \mathbb{R} \to \,]-\pi/2, \pi/2[}. $$ For \(x \in \mathbb{R}\), \(\arctan(x)\) is the unique \(\theta \in \,]-\pi/2, \pi/2[\) such that \(\tan \theta = x\).
Proposition — Properties of \(\arccos\)\(\virgule\) \(\arcsin\)\(\virgule\) \(\arctan\)
  • \(\arccos\) is continuous on \([-1, 1]\), of class \(C^\infty\) on \(]-1, 1[\), with $$ \textcolor{colorprop}{\forall x \in \,]-1, 1[, \quad \arccos'(x) = -\frac{1}{\sqrt{1 - x^2}}.} $$
  • \(\arcsin\) is continuous on \([-1, 1]\), of class \(C^\infty\) on \(]-1, 1[\), odd, with $$ \textcolor{colorprop}{\forall x \in \,]-1, 1[, \quad \arcsin'(x) = \frac{1}{\sqrt{1 - x^2}}.} $$
  • \(\arctan\) is of class \(C^\infty\) on \(\mathbb{R}\), odd, with $$ \textcolor{colorprop}{\forall x \in \mathbb{R}, \quad \arctan'(x) = \frac{1}{1 + x^2}.} $$
Moreover, \(\arccos\) is strictly decreasing on \([-1, 1]\), while \(\arcsin\) and \(\arctan\) are strictly increasing on their domains. At the endpoints \(\pm 1\), \(\arccos\) and \(\arcsin\) have no finite one-sided derivative; geometrically, their graphs have vertical tangents there. The function \(\arctan\) has horizontal asymptotes \(y = \pm \pi/2\) at \(\pm\infty\).

  • \(\arctan'\). The inverse-function differentiation theorem of Real functions applies to \(\tan\) on \(]-\pi/2, \pi/2[\), where \(\tan' = 1 + \tan^2\) is non-zero. So \(\arctan\) is differentiable on \(\mathbb{R}\) and, with \(\theta = \arctan x\), that is, \(\tan \theta = x\): $$ \arctan'(x) = \frac{1}{\tan'(\theta)} = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + x^2}. $$ This holds for every \(x \in \mathbb{R}\). Iterating, \(\arctan\) is of class \(C^\infty\).
  • \(\arcsin'\) on \(]-1, 1[\). Apply the same theorem to \(\sin\) on \(]-\pi/2, \pi/2[\), where \(\sin' = \cos\) is non-zero (the endpoints \(\pm \pi/2\) are excluded). So \(\arcsin\) is differentiable on \(]-1, 1[\) (the image of the open interval). With \(\theta = \arcsin x\), that is, \(\sin \theta = x\) and \(\theta \in \,]-\pi/2, \pi/2[\), \(\cos \theta > 0\), so \(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}\) (positive square root), and $$ \arcsin'(x) = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1 - x^2}}. $$
  • \(\arccos'\) on \(]-1, 1[\). Same technique with \(\cos\) on \(]0, \pi[\), where \(\cos' = -\sin\) is non-zero. With \(\theta = \arccos x\), that is, \(\cos \theta = x\) and \(\theta \in \,]0, \pi[\), \(\sin \theta > 0\), so \(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}\), and $$ \arccos'(x) = \frac{1}{-\sin \theta} = -\frac{1}{\sqrt{1 - x^2}}. $$
  • Continuity on the closed interval \([-1, 1]\). The functions \(\sin|_{[-\pi/2, \pi/2]}\) and \(\cos|_{[0, \pi]}\) are continuous strictly monotone bijections from a closed interval onto a closed interval. By the « continuous strictly monotone bijection » theorem of Real functions, their inverses are continuous on the closed image \([-1, 1]\). Hence \(\arccos\) and \(\arcsin\) are continuous on \([-1, 1]\) (including at \(\pm 1\)).
  • Endpoint behaviour at \(\pm 1\). At \(x = 1\): \(\arcsin(1) = \pi/2\), and the corresponding value of \(\sin'\) is \(\cos(\pi/2) = 0\), so the inverse-function formula breaks down at the endpoint. Concretely, the derivative formula \(\arcsin'(x) = 1/\sqrt{1 - x^2}\) on \(]-1, 1[\) tends to \(+\infty\) in absolute value as \(x \to \pm 1\), which corresponds to a vertical tangent. Hence \(\arcsin\) has no finite one-sided derivative at \(\pm 1\). The same argument applies at \(x = \pm 1\) for \(\arccos\).
  • Parity. \(\arcsin\) is odd: if \(\theta = \arcsin x\), then \(\sin(-\theta) = -\sin \theta = -x\) and \(-\theta \in \,[-\pi/2, \pi/2]\), so \(\arcsin(-x) = -\theta = -\arcsin(x)\). Same argument for \(\arctan\) on \(\mathbb{R}\). (\(\arccos\) is not odd, because its image \([0, \pi]\) is not symmetric about \(0\); the relation is \(\arccos(-x) = \pi - \arccos(x)\), an exercise.)
  • Asymptotes of \(\arctan\). As \(x \to +\infty\), \(\arctan x \to \pi/2\) (since \(\tan \theta \to +\infty\) as \(\theta \to (\pi/2)^-\)). By oddness, \(\arctan x \to -\pi/2\) at \(-\infty\). The horizontal lines \(y = \pm \pi/2\) are asymptotes.

Proposition — Useful identities
  • For every \(x \in [-1, 1]\): $$ \textcolor{colorprop}{\arccos(x) + \arcsin(x) = \frac{\pi}{2}.} $$
  • For \(x > 0\): \(\textcolor{colorprop}{\arctan(x) + \arctan(1/x) = \pi/2}\). For \(x < 0\): \(\textcolor{colorprop}{\arctan(x) + \arctan(1/x) = -\pi/2}\).
  • For every \(x \in [-1, 1]\): $$ \textcolor{colorprop}{\cos(\arcsin x) = \sin(\arccos x) = \sqrt{1 - x^2}.} $$

We prove the first identity in detail; the rest are similar.
  • Step 1 --- derivative argument on the open interval \(]-1, 1[\). Set \(\varphi(x) = \arccos(x) + \arcsin(x)\) for \(x \in [-1, 1]\). The function \(\varphi\) is differentiable on \(]-1, 1[\), where the derivative formulas of P5.1 apply, with $$ \varphi'(x) = -\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x^2}} = 0. $$ So \(\varphi\) is constant on \(]-1, 1[\).
  • Step 2 --- value of the constant. Evaluate at \(x = 0\): \(\arccos(0) = \pi/2\) (since \(\cos(\pi/2) = 0\) and \(\pi/2 \in [0, \pi]\)), and \(\arcsin(0) = 0\). So \(\varphi(0) = \pi/2 + 0 = \pi/2\). Hence \(\varphi(x) = \pi/2\) for every \(x \in \,]-1, 1[\).
  • Step 3 --- boundary case \(x = \pm 1\). The function \(\varphi\) is continuous on the closed interval \([-1, 1]\) (P5.1: \(\arccos\) and \(\arcsin\) are both continuous on \([-1, 1]\)). Since \(\varphi(x) = \pi/2\) for \(x \in \,]-1, 1[\), by continuity \(\varphi(\pm 1) = \pi/2\) as well. Direct verification: \(\varphi(1) = \arccos(1) + \arcsin(1) = 0 + \pi/2 = \pi/2\) \(\checkmark\), and \(\varphi(-1) = \arccos(-1) + \arcsin(-1) = \pi + (-\pi/2) = \pi/2\) \(\checkmark\).
The identity \(\cos(\arcsin x) = \sqrt{1 - x^2}\) follows from \(\cos^2 \theta + \sin^2 \theta = 1\) with \(\theta = \arcsin x\): \(\cos^2 \theta = 1 - x^2\), and since \(\theta \in [-\pi/2, \pi/2]\), \(\cos \theta \ge 0\), hence \(\cos \theta = \sqrt{1 - x^2}\). The other identity \(\sin(\arccos x) = \sqrt{1 - x^2}\) follows from \(\arccos x + \arcsin x = \pi/2\): \(\arccos x = \pi/2 - \arcsin x\), so \(\sin(\arccos x) = \sin(\pi/2 - \arcsin x) = \cos(\arcsin x) = \sqrt{1 - x^2}\). The \(\arctan\) identities are obtained similarly: for \(x > 0\), set \(\psi(x) = \arctan x + \arctan(1/x)\); \(\psi'(x) = 1/(1 + x^2) + (-1/x^2)/(1 + 1/x^2) = 1/(1+x^2) - 1/(x^2 + 1) = 0\), so \(\psi\) is constant on \(]0, +\infty[\); \(\psi(1) = 2 \arctan 1 = 2 \cdot \pi/4 = \pi/2\). For \(x < 0\), \(-x > 0\), so \(\psi(-x) = \pi/2\); by odd parity of \(\arctan\) (and hence of \(\psi\)), \(\psi(x) = -\psi(-x) = -\pi/2\).

Example
Compute \(\arctan(1) + \arctan(2) + \arctan(3)\).

Set \(S = \arctan 1 + \arctan 2 + \arctan 3\). By the addition formula for \(\tan\) (and using \(\arctan 2 + \arctan 3\)): $$ \tan(\arctan 2 + \arctan 3) = \frac{2 + 3}{1 - 2 \cdot 3} = \frac{5}{-5} = -1. $$ Since \(\arctan 2 \in \,]\pi/4, \pi/2[\) and \(\arctan 3 \in \,]\pi/4, \pi/2[\), the sum \(\arctan 2 + \arctan 3 \in \,]\pi/2, \pi[\), an interval where \(\tan = -1\) is reached at \(3\pi/4\). So \(\arctan 2 + \arctan 3 = 3\pi/4\). Adding \(\arctan 1 = \pi/4\): \(S = \pi/4 + 3\pi/4 = \pi\).

Example
For \(x \in \mathbb{R}\), simplify \(\sin(\arctan x)\).

Set \(\theta = \arctan x \in \,]-\pi/2, \pi/2[\), so \(\tan \theta = x\) and \(\cos \theta > 0\). Using \(\sin \theta = \tan \theta \cdot \cos \theta\) and \(\cos \theta = 1/\sqrt{1 + \tan^2 \theta} = 1/\sqrt{1 + x^2}\): $$ \sin(\arctan x) = \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} = \frac{x}{\sqrt{1 + x^2}}. $$

Example
Solve on \([-1, 1]\): \(\arccos(x) = \arcsin(x)\).

Use P5.2: \(\arccos x + \arcsin x = \pi/2\). The equation \(\arccos x = \arcsin x\) then gives \(2 \arcsin x = \pi/2\), hence \(\arcsin x = \pi/4\), hence \(x = \sin(\pi/4) = \sqrt{2}/2\). Verification: \(\arccos(\sqrt{2}/2) = \pi/4 = \arcsin(\sqrt{2}/2)\) \(\checkmark\).

Example — Domain trap
Compute \(\arccos(\cos(5\pi/4))\). (Beware: the answer is not \(5\pi/4\).)

The function \(\arccos\) takes values in \([0, \pi]\), so the answer must lie in \([0, \pi]\) --- and \(5\pi/4 \notin [0, \pi]\). Compute first \(\cos(5\pi/4)\): \(5\pi/4 = \pi + \pi/4\), so \(\cos(5\pi/4) = -\cos(\pi/4) = -\sqrt{2}/2\). Now apply \(\arccos\): $$ \arccos(\cos(5\pi/4)) = \arccos(-\sqrt{2}/2). $$ The unique \(\theta \in [0, \pi]\) with \(\cos \theta = -\sqrt{2}/2\) is \(\theta = 3\pi/4\). So \(\arccos(\cos(5\pi/4)) = 3\pi/4\). Moral. The identity \(\arccos(\cos \theta) = \theta\) holds only for \(\theta \in [0, \pi]\); outside, you have to first reduce \(\theta\) modulo \(2\pi\) and then use the parity / phase symmetries of \(\cos\) to bring the angle back into \([0, \pi]\).

Example — Graphs of \(\arccos\)\(\virgule\) \(\arcsin\)\(\virgule\) \(\arctan\)
\(\arccos\) goes from \(\pi\) to \(0\) as \(x\) goes from \(-1\) to \(1\) (decreasing). \(\arcsin\) goes from \(-\pi/2\) to \(\pi/2\) (increasing, odd, vertical tangent at \(\pm 1\)). \(\arctan\) is bounded on \(\mathbb{R}\) by the asymptotes \(y = \pm \pi/2\) (increasing, odd).
Method — Simplifying \(\arctan a + \arctan b\)
For \(a, b \in \mathbb{R}\) with \(ab \ne 1\), the addition formula for \(\tan\) gives $$ \tan(\arctan a + \arctan b) = \frac{a + b}{1 - ab}, $$ so \(\arctan a + \arctan b\) equals \(\arctan\!\bigl( \frac{a+b}{1 - ab} \bigr) + k\pi\) for some integer \(k \in \{-1, 0, 1\}\). The integer \(k\) is fixed by checking the interval into which \(\arctan a + \arctan b\) falls:
  • \(k = 0\) if the sum is in \(\,]-\pi/2, \pi/2[\);
  • \(k = 1\) if it is in \(\,]\pi/2, \pi[\) (typically when \(a, b > 0\) and \(ab > 1\));
  • \(k = -1\) if it is in \(\,]-\pi, -\pi/2[\) (typically when \(a, b < 0\) and \(ab > 1\)).
The case \(ab = 1\) is the boundary: \(\arctan a + \arctan(1/a) = \pm \pi/2\) depending on the sign of \(a\) (cf. P5.2).
Skills to practice
  • Computing values and exact expressions
  • Simplifying \(\arctan a + \arctan b\)
  • Solving equations involving inverse trig
VI Complex-valued functions of a real variable
A function \(f : I \to \mathbb{C}\) defined on a real interval \(I \subset \mathbb{R}\) decomposes into real and imaginary parts \(f = u + iv\) with \(u, v : I \to \mathbb{R}\). Differentiability and the derivative are defined component-wise. The exponential \(\exp(g)\) for \(g : I \to \mathbb{C}\) derivable is the only specifically-complex case the program demands; the formula \((\exp g)' = g' \, \exp g\) extends naturally from the real case and is constantly used in linear differential equations with complex coefficients.
Definition — Complex-valued real-variable function\(\virgule\) derivative
Let \(I \subset \mathbb{R}\) be an interval and \(f : I \to \mathbb{C}\). Set \(u = \mathrm{Re}(f) : I \to \mathbb{R}\) and \(v = \mathrm{Im}(f) : I \to \mathbb{R}\), so that \(f = u + iv\). For \(a \in I\) (interior point, or one-sided derivative if \(a\) is an endpoint), we say \(f\) is differentiable at \(a\) when both \(u\) and \(v\) are differentiable at \(a\), and define $$ \textcolor{colordef}{f'(a) = u'(a) + i \, v'(a).} $$ We say \(f\) is differentiable on \(I\) if \(u\) and \(v\) are; the derivative is the function \(f' = u' + i v' : I \to \mathbb{C}\). Class \(C^k\) on \(I\) is defined the same way: \(f \in C^k(I, \mathbb{C})\) iff \(u, v \in C^k(I, \mathbb{R})\).
Proposition — Operations on complex-valued differentiable functions
Let \(f, g : I \to \mathbb{C}\) differentiable on \(I\), \(\lambda, \mu \in \mathbb{C}\). Then \(\lambda f + \mu g\), \(fg\) are differentiable on \(I\), and if \(g\) does not vanish on \(I\), \(f/g\) is differentiable on \(I\). The derivatives are $$ \textcolor{colorprop}{(\lambda f + \mu g)' = \lambda f' + \mu g', \qquad (fg)' = f' g + f g', \qquad \left( \frac{f}{g} \right)' = \frac{f' g - f g'}{g^2}.} $$ (Proof component-wise from the real case via \(f = u + iv\) and \(g = U + iV\) with \(u, v, U, V : I \to \mathbb{R}\); admitted.)
Proposition — Derivative of \(\exp(g)\) for \(g\) complex-valued
Let \(I \subset \mathbb{R}\) be an interval and \(g : I \to \mathbb{C}\) differentiable on \(I\). Define \(\exp(g) : I \to \mathbb{C}\) by \(\exp(g)(x) = \exp(g(x))\), where the right-hand side uses the complex exponential from Complex numbers: \(\exp(a + ib) = \mathrm{e}^a (\cos b + i \sin b)\). Then \(\exp(g)\) is differentiable on \(I\), and $$ \textcolor{colorprop}{(\exp(g))'(x) = g'(x) \cdot \exp(g(x)).} $$

Write \(g = u + iv\) with \(u, v : I \to \mathbb{R}\) differentiable. Using the complex exponential formula: $$ \exp(g(x)) = \mathrm{e}^{u(x)} \cos(v(x)) + i \, \mathrm{e}^{u(x)} \sin(v(x)) = U(x) + i V(x), $$ with \(U(x) = \mathrm{e}^{u(x)} \cos(v(x))\) and \(V(x) = \mathrm{e}^{u(x)} \sin(v(x))\). By the real product/chain rules: $$ \begin{aligned} U'(x) &= u'(x) \, \mathrm{e}^{u(x)} \cos(v(x)) - v'(x) \, \mathrm{e}^{u(x)} \sin(v(x)), \\ V'(x) &= u'(x) \, \mathrm{e}^{u(x)} \sin(v(x)) + v'(x) \, \mathrm{e}^{u(x)} \cos(v(x)). \end{aligned} $$ Hence (factoring \(\mathrm{e}^{u(x)}\)): $$ (\exp(g))'(x) = U'(x) + i V'(x) = \mathrm{e}^{u(x)} \bigl[ (u' \cos v - v' \sin v) + i (u' \sin v + v' \cos v) \bigr]. $$ Factorization step. Expand the candidate factor \((u' + iv')(\cos v + i \sin v)\): $$ \begin{aligned} (u' + iv')(\cos v + i \sin v) &= u' \cos v + i u' \sin v + i v' \cos v + i^2 v' \sin v \\ &= (u' \cos v - v' \sin v) + i (u' \sin v + v' \cos v). \end{aligned} $$ This matches the bracket exactly. So $$ \begin{aligned} (\exp(g))'(x) &= \mathrm{e}^{u(x)} (u'(x) + i v'(x))(\cos(v(x)) + i \sin(v(x))) && \text{(factorization step)} \\ &= (u'(x) + i v'(x)) \cdot \mathrm{e}^{u(x)} (\cos(v(x)) + i \sin(v(x))) && \text{(commute scalar \(\mathrm{e}^u\))} \\ &= g'(x) \cdot \exp(g(x)). && \text{(definitions of \(g'\) and \(\exp(g)\))} \end{aligned} $$

Example
For \(\omega \in \mathbb{R}\), compute the derivative of \(f : x \mapsto \mathrm{e}^{i \omega x}\) on \(\mathbb{R}\).

Apply P6.2 with \(g(x) = i \omega x\), \(g'(x) = i \omega\): $$ f'(x) = (\exp(g))'(x) = g'(x) \exp(g(x)) = i \omega \, \mathrm{e}^{i \omega x}. $$

Example
Express \(\cos(\omega x)\) and \(\sin(\omega x)\) as the real and imaginary parts of \(\mathrm{e}^{i \omega x}\), and recover the derivative formulas \((\cos \omega x)' = -\omega \sin \omega x\), \((\sin \omega x)' = \omega \cos \omega x\).

By definition: \(\mathrm{e}^{i \omega x} = \cos(\omega x) + i \sin(\omega x)\). Differentiating using Ex6.1: \((\mathrm{e}^{i \omega x})' = i \omega \, \mathrm{e}^{i \omega x} = i \omega (\cos \omega x + i \sin \omega x) = -\omega \sin \omega x + i \omega \cos \omega x\). Identifying real and imaginary parts on each side: \((\cos \omega x)' = -\omega \sin \omega x\), \((\sin \omega x)' = \omega \cos \omega x\) \(\checkmark\).

Example
For \(n \in \mathbb{N}^*\), compute the derivative of \(f : x \mapsto (1 + ix)^n\) on \(\mathbb{R}\).

The function is a polynomial in \(x\) with complex coefficients, hence differentiable. By induction or by the product rule (P6.1): $$ f'(x) = n \cdot (i) \cdot (1 + ix)^{n-1} = i \, n \, (1 + ix)^{n - 1}. $$ (Base case \(n = 1\): \((1 + ix)' = i\). Heredity: assume true at rank \(n\); then \(((1+ix)^{n+1})' = ((1+ix)^n \cdot (1+ix))' = i n (1+ix)^{n-1} \cdot (1+ix) + (1+ix)^n \cdot i = i(n+1)(1+ix)^n\).)

Method — Derivatives of trigonometric expressions via \(\mathrm{e}^{ix}\)
When an expression mixes \(\cos\) and \(\sin\) with arguments \(\omega x\) (or any linear combination), the cleanest path is to write $$ \cos(\omega x) = \mathrm{Re}(\mathrm{e}^{i \omega x}), \qquad \sin(\omega x) = \mathrm{Im}(\mathrm{e}^{i \omega x}), $$ manipulate the complex exponential algebraically (it is much easier to differentiate and to combine), then take real or imaginary parts at the end. The same trick handles primitives, linearisation, and addition formulas. Example. The product-to-sum formula \(\cos(\omega x) \cos(\omega' x) = \tfrac{1}{2} \bigl( \cos((\omega + \omega') x) + \cos((\omega - \omega') x) \bigr)\) is recovered from \(\mathrm{Re}(\mathrm{e}^{i \omega x}) \cdot \mathrm{Re}(\mathrm{e}^{i \omega' x}) = \tfrac{1}{4}(\mathrm{e}^{i\omega x} + \mathrm{e}^{-i\omega x})(\mathrm{e}^{i\omega' x} + \mathrm{e}^{-i\omega' x})\) by expanding and pairing \(\mathrm{e}^{i k x} + \mathrm{e}^{-i k x} = 2 \cos(kx)\).
Skills to practice
  • Differentiating complex-valued real-variable functions