Revision: Functions and Differentiation

For all strictly positive real numbers \(a\) and \(b\):$$\ln(a \times b) = \ln(a) + \ln(b)$$

For all \(a > 0\), \(b > 0\) and any integer \(n\):

• Inverse: \(\ln\left(\frac{1}{b}\right) = -\ln(b)\)
• Quotient: \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• Power: \(\ln(a^n) = n\ln(a)\)
• Square Root: \(\ln(\sqrt{a}) = \frac{1}{2}\ln(a)\)

Let \(u\) be a differentiable and strictly positive function on an interval \(I\). The function \(f(x) = \ln(u(x))\) is differentiable on \(I\) and:$$\textcolor{colorprop}{(\ln(u(x)))' = \frac{u'(x)}{u(x)}}$$

This is the method for a statement whose two sides each reduce to a single logarithm. It does not cover statements such as \((\ln x)^2 + \ln x = 0\), where \(\ln x\) appears squared: those call for the substitution \(X = \ln x\), which turns them into an ordinary equation in \(X\).

1. Domain first. Every \(\ln\) in the statement demands its argument strictly positive. One solves those conditions together and keeps their intersection: outside it, the statement has no meaning, and no solution can come from there.
2. One logarithm on each side. Using the laws of logarithms, one rewrites the statement as \(\ln A = \ln B\), or as \(\ln A \le \ln B\).
3. Remove the logarithms. \(\ln\) is strictly increasing on \((0, +\infty)\), so it preserves both equality and order. For \(A > 0\) and \(B > 0\) — which step 1 has secured: $$\ln A = \ln B \iff A = B \qquad \text{and} \qquad \ln A \le \ln B \iff A \le B.$$
4. Intersect with the domain. A value found at step 3 is a solution only if it lies in the domain of step 1.

Step 4 is not a formality, and it is worth being exact about why. Removing the logarithms is a true equivalence on the domain — nothing is lost and nothing is gained there. What widens the search is the next move: the equation obtained is then solved on all of \(\mathbb{R}\), where it may well have roots the original statement never had. Those candidates are not errors of the method; they are values the domain must reject.

The derivative of the natural exponential function is itself:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}(e^x) = e^x $$

The derivative of an exponential function with a function in the exponent is given by:$$ \left(e^{u(x)}\right)' = e^{u(x)}u'(x) $$ In Leibniz notation:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{u}\right) = e^{u}\dfrac{\mathrm{d}u}{\mathrm{d}x} $$

If \(f(x)=\dfrac{u(x)}{v(x)}\) with \(u\) and \(v\) differentiable and \(v(x) \neq 0\), then$$ f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$ In Leibniz notation, \(y=\dfrac{u}{v}\), then$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} $$

For a function \(f\) that is twice differentiable on an interval \(I\):

• \(f''(x) \ge 0\) for all \(x \in I\), if and only if \(f\) is concave up on \(I\).
• \(f''(x) \le 0\) for all \(x \in I\), if and only if \(f\) is concave down on \(I\).

Let \(f\) be a function twice differentiable on an open interval containing \(a\), with \(f'(a) = 0\).

• If \(f''(a) > 0\), then \(f\) has a local minimum at \(a\).
• If \(f''(a) < 0\), then \(f\) has a local maximum at \(a\).
• If \(f''(a) = 0\), the test is inconclusive: \(f\) may have a local minimum, a local maximum or neither at \(a\), and another argument is needed, such as the sign of \(f'\) on either side of \(a\).

A point \((a, f(a))\) is a point of inflexion if \(f''(a)=0\) and the sign of \(f''(x)\) changes at \(x=a\).
The point of inflexion is:

• stationary if \(f'(a)=0\).
• non-stationary if \(f'(a) \neq 0\).

For \(x\) in radians, the derivatives of the basic trigonometric functions are:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}(\sin x) = \cos x \quad \text{and} \quad \dfrac{\mathrm{d}}{\mathrm{d}x}(\cos x) = -\sin x $$

For a differentiable function \(u\):

• \((\sin(u(x)))' = u'(x)\cos(u(x))\)
• \((\cos(u(x)))' = -u'(x)\sin(u(x))\)
• \((\tan(u(x)))' = u'(x)\bigl(1+\tan^2(u(x))\bigr)\), where \(\cos(u(x)) \neq 0\)

In Leibniz notation:

• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin(u)) = \cos(u)\dfrac{\mathrm{d}u}{\mathrm{d}x}\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\cos(u)) = -\sin(u)\dfrac{\mathrm{d}u}{\mathrm{d}x}\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\tan(u)) = \bigl(1+\tan^2(u)\bigr)\dfrac{\mathrm{d}u}{\mathrm{d}x}\)