International Baccalaureate · DP 2: Applications and Interpretation HL

Rules of Differentiation

⌚ ~154 min ▢ 17 blocks ✓ 84 exercises ➣ Prerequisites : Introduction to Differential Calculus, Functions, Exponentials, Logarithms, Trigonometric Functions

I Products, Quotients and Composites

I.1 Product Rule

While the derivative of a sum is the sum of the derivatives, the same is not true for a product. To find the derivative of a function that is the product of two other functions, such as $f(x) = (x^2+1)\sqrt{x}$, we must use a specific formula called the Product Rule.

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Proposition — Product Rule

If $f(x)=u(x)v(x)$ with $u$ and $v$ differentiable, then$$ f'(x) = u'(x)v(x) + u(x)v'(x) $$ In Leibniz notation, if $y=u\cdot v$ :$$ \dfrac{\mathrm{d}y}{\mathrm{d}x}= \dfrac{\mathrm{d}u}{\mathrm{d}x}v + u\dfrac{\mathrm{d}v}{\mathrm{d}x} $$

In words: “The derivative of the first times the second, plus the first times the derivative of the second.”

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Example

Find the derivative of $f(x) = (x+1)(x^2+3)$.

Answer

Using prime notation:
For $f(x)=u(x)v(x)$ where $u(x)=x+1$ and $v(x)=x^2+3$, the derivatives are $u'(x)=1$ and $v'(x)=2x$.$$\begin{aligned}f'(x) &= u'(x)v(x) + u(x)v'(x) \\ &= (1)(x^2+3) + (x+1)(2x) \\ &= x^2+3 + 2x^2+2x \\ &= 3x^2+2x+3\end{aligned}$$
Using Leibniz's notation ($y=f(x)$):
For $y=uv$ where $u=x+1$ and $v=x^2+3$, the derivatives are $\frac{\mathrm{d}u}{\mathrm{d}x}=1$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=2x$.$$\begin{aligned}\dfrac{\mathrm{d}y}{\mathrm{d}x} &= \dfrac{\mathrm{d}u}{\mathrm{d}x}v + u\dfrac{\mathrm{d}v}{\mathrm{d}x} \\ &= (1)(x^2+3) + (x+1)(2x) \\ &= x^2+3 + 2x^2+2x \\ &= 3x^2+2x+3\end{aligned}$$

Skills to practice

Differentiating with the Product Rule

I.2 Quotient Rule

Just as with products, finding the derivative of a quotient of two functions requires a specific formula. The Quotient Rule is used to differentiate functions of the form $f(x) = \dfrac{u(x)}{v(x)}$, such as $f(x) = \dfrac{2x+1}{x^2+1}$.

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Proposition — Quotient Rule

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$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} $$

Example

Find the derivative of $f(x) = \frac{x}{x+1}$.

Answer

Using prime notation:
For $f(x)=\frac{u(x)}{v(x)}$ where $u(x)=x$ and $v(x)=x+1$, the derivatives are $u'(x)=1$ and $v'(x)=1$.$$\begin{aligned}f'(x) &= \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \\ &= \frac{(1)(x+1) - (x)(1)}{(x+1)^2} \\ &= \frac{x+1-x}{(x+1)^2} \\ &= \frac{1}{(x+1)^2}\end{aligned}$$
Using Leibniz's notation $(y=f(x))$:
For $y=\frac{u}{v}$ where $u=x$ and $v=x+1$, the derivatives are $\frac{\mathrm{d}u}{\mathrm{d}x}=1$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=1$.$$\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \\ &= \frac{(1)(x+1) - (x)(1)}{(x+1)^2} \\ &= \frac{x+1-x}{(x+1)^2} \\ &= \frac{1}{(x+1)^2}\end{aligned}$$

Skills to practice

Differentiating with the Quotient Rule

I.3 Derivative of $x \mapsto g(ax+b)$

Many functions are obtained by plugging a linear expression $ax+b$ into a reference function $g$: $\sqrt{2x+1}$ (with $g(X) = \sqrt{X}$), $(3x-1)^2$ (with $g(X) = X^2$), $\dfrac{1}{4x+2}$ (with $g(X) = \dfrac{1}{X}$). Their derivative follows from the derivative of $g$ by a simple rule.

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Proposition — Derivative of $x \mapsto g(ax+b)$

Let $g$ be a function differentiable on an interval $J$, and $a$, $b$ two real numbers. The function $f$ defined by $f(x) = g(ax+b)$ is differentiable at every $x$ such that $ax+b \in J$, and $$ f'(x) = a \times g'(ax+b). $$

Example

Find the derivative of each function.

$f(x) = (3x-1)^2$ for all real $x$.
$f(x) = \sqrt{2x+1}$ for $x > -\dfrac{1}{2}$.
$f(x) = \dfrac{1}{4x+2}$ for $x \neq -\dfrac{1}{2}$.

Answer

$f(x) = g(3x-1)$ with $g(X) = X^2$, $g'(X) = 2X$ and $a = 3$: $$ f'(x) = 3 \times 2(3x-1) = 6(3x-1) = 18x - 6. $$ Check: expanding, $f(x) = 9x^2 - 6x + 1$, whose derivative is indeed $18x - 6$.
$f(x) = g(2x+1)$ with $g(X) = \sqrt{X}$, $g'(X) = \dfrac{1}{2\sqrt{X}}$ (for $X > 0$) and $a = 2$: $$ f'(x) = 2 \times \dfrac{1}{2\sqrt{2x+1}} = \dfrac{1}{\sqrt{2x+1}}. $$
$f(x) = g(4x+2)$ with $g(X) = \dfrac{1}{X}$, $g'(X) = -\dfrac{1}{X^2}$ (for $X \neq 0$) and $a = 4$: $$ f'(x) = 4 \times \left(-\dfrac{1}{(4x+2)^2}\right) = -\dfrac{4}{(4x+2)^2}. $$

This rule is a special case of the chain rule, which gives the derivative of $g(u(x))$ for any differentiable function $u$, not only a linear one $ax+b$.

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Skills to practice

Differentiating $g(ax+b)$ and Its Products
Writing Up Derivatives with the Rules

I.4 Chain Rule

Many complex functions are created by composing simpler ones. A composite function has the form $f(x) = v(u(x))$, where one function (the “inner” function, $u$) is the input to another (the “outer” function, $v$). For example, in the function $f(x) = (x^2+1)^3$, the inner function is $u(x)=x^2+1$ and the outer function is $v(x)=x^3$.
The Chain Rule provides a powerful method for differentiating such functions by finding the derivatives of the inner and outer functions separately.

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Proposition — Chain Rule

If $u$ is differentiable at $x$ and $v$ is differentiable at $u(x)$, then the function $f$ defined by $f(x)=v(u(x))$ is differentiable at $x$ and:$$ f'(x) = v'(u(x)) \cdot u'(x) $$ In Leibniz notation, if $y=v(u)$ then$$ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \cdot \dfrac{\mathrm{d}u}{\mathrm{d}x} $$

Example

Find the derivative of $f(x) = (x^2+1)^3$.

Answer

Using prime notation:
Let the outer function be $v(x)=x^3$ and the inner function be $u(x)=x^2+1$. We have $f(x)=v(u(x))$. The derivatives are $v'(x)=3x^2$ and $u'(x)=2x$.$$\begin{aligned}f'(x) &= v'(u(x)) \cdot u'(x) \\ &= 3(u(x))^2 \cdot (2x) \\ &= 3(x^2+1)^2 \cdot (2x)\\ &= 6x(x^2+1)^2\end{aligned}$$
Using Leibniz's notation ($y=f(x)$):
For $y=u^3$ and $u=x^2+1$, the derivatives are $\frac{\mathrm{d}y}{\mathrm{d}u}=3u^2$ and $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$.$$\begin{aligned}\dfrac{\mathrm{d}y}{\mathrm{d}x} &= \dfrac{\mathrm{d}y}{\mathrm{d}u} \cdot \dfrac{\mathrm{d}u}{\mathrm{d}x} \\ &= (3u^2) \cdot (2x) \\ &= 3(x^2+1)^2 \cdot (2x)\\ & = 6x(x^2+1)^2\end{aligned}$$

Skills to practice

Choosing a Differentiation Rule
Decomposing Composite Functions
Differentiating with the Chain Rule

II Derivatives of Standard Functions

II.1 Exponential Functions

The exponential function, particularly $f(x)=e^x$, holds a unique and fundamental place in calculus. Its rate of change is directly proportional to its value, which makes it the mathematical language of many natural processes like population growth and compound interest. This function has a remarkably simple derivative.

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Proposition — Derivative of $e^x$

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

Proposition — Chain Rule for $e^{u(x)}$

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:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{u}\right) = e^{u}\dfrac{\mathrm{d}u}{\mathrm{d}x} $$

Example

Find the derivative of $f(x)=e^{x^2}$.

Answer

Here, the inner function is $u(x)=x^2$, so $u'(x)=2x$.$$\begin{aligned}f'(x) &= e^{u(x)}u'(x)\\ &= e^{x^2}2x\end{aligned} $$

Skills to practice

Differentiating Composite Exponential Functions
Applying the Product and Quotient Rules to the Exponential

II.2 Logarithmic Functions

The natural logarithm function, $f(x)=\ln(x)$, is the inverse of the exponential function $e^x$. This inverse relationship is the key to finding its derivative. Just as the exponential function has a remarkably simple derivative, so does the logarithm. This derivative is fundamental in calculus, particularly in integration, as it gives an antiderivative of $x \mapsto 1/x$ on $(0, +\infty)$.

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Proposition — Derivative of $\ln x$

The derivative of the natural logarithm function for $x>0$ is:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}(\ln x) = \dfrac{1}{x} $$

Proposition — Chain Rule for $\ln(u(x))$

If $u$ is differentiable and strictly positive on an interval $I$, then $\ln(u)$ is differentiable on $I$ and:$$ \left(\ln(u(x))\right)' = \dfrac{u'(x)}{u(x)} $$ In Leibniz notation:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(u)\right) = \dfrac{1}{u}\,\dfrac{\mathrm{d}u}{\mathrm{d}x} $$

Example

Find the derivative of $f(x) = \ln(2x+3)$.

Answer

This is of the form $\ln(u(x))$ where the inner function is $u(x)=2x+3$.
The derivative of the inner function is $u'(x)=2$.
Using the chain rule for logarithms:$$ f'(x) = \dfrac{u'(x)}{u(x)} = \dfrac{2}{2x+3} $$

Method — Simplify Before Differentiating

For a complicated logarithmic function, the laws of logarithms are used to expand or simplify the expression before differentiating, on an interval where the arguments of the logarithms are positive. This can turn a difficult problem into a very simple one.
Recall the laws, for $a>0$ and $b>0$: $\ln(ab)=\ln(a)+\ln(b)$, $\ln(a/b)=\ln(a)-\ln(b)$, and $\ln(a^n)=n\ln(a)$.

Example

Find the derivative of $f(x) = \ln\left(\dfrac{x+1}{x}\right)$.

Answer

$f$ is defined where $\dfrac{x+1}{x} > 0$, that is for $x < -1$ or $x > 0$. For $x > 0$, both $x+1$ and $x$ are positive, so the law $\ln(a/b)=\ln(a)-\ln(b)$ simplifies the expression:$$ f(x) = \ln(x+1) - \ln(x) $$ Now, we can differentiate this much simpler expression term by term:$$\begin{aligned}f'(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x}(\ln(x+1)) - \dfrac{\mathrm{d}}{\mathrm{d}x}(\ln x) \\ &= \dfrac{1}{x+1} - \dfrac{1}{x}\end{aligned}$$ For $x < -1$, the rule for $\ln(u)$ with $u = \dfrac{x+1}{x}$ gives the same expression of $f'(x)$.

Skills to practice

Differentiating Logarithmic Functions

II.3 Trigonometric Functions

Trigonometric functions describe periodic phenomena, which are common in physics and engineering. Their derivatives follow a cyclical pattern that is both elegant and powerful. The derivatives of $\sin(x)$ and $\cos(x)$ come from the limit definition of the derivative, with $x$ measured in radians.

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Proposition — Derivatives of Sine and Cosine

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

Proposition — Derivative of Tangent

The derivative of the tangent function is:$$ \dfrac{\mathrm{d}}{\mathrm{d}x}(\tan x) = 1+\tan^2 x $$

Proof

We use the quotient rule on $f(x) = \tan x = \dfrac{\sin x}{\cos x}$.
Let $u=\sin x$ and $v=\cos x$. Then $u'=\cos x$ and $v'=-\sin x$.$$\begin{aligned}f'(x) &= \dfrac{u'v - uv'}{v^2}\\ &= \dfrac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} \\ &= \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}\\ &= \dfrac{\cos^2 x}{\cos^2 x} + \dfrac{\sin^2 x}{\cos^2 x} \\ &= 1+ \dfrac{\sin^2 x}{\cos^2 x}\\ &= 1+\tan^2 x \\ \end{aligned}$$

Proposition — Chain Rule for Trigonometric Functions

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:

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

Example

Find the derivative of $f(x) = \cos(x^2+3x)$.

Answer

This is of the form $\cos(u(x))$ where the inner function is $u(x)=x^2+3x$.
The derivative of the inner function is $u'(x)=2x+3$.
Using the chain rule for cosine:$$\begin{aligned}f'(x) &= -u'(x)\sin(u(x)) \\ &= -(2x+3)\sin(x^2+3x)\end{aligned}$$

Skills to practice

Recognising Derivatives of Standard Functions
Differentiating Trigonometric Functions
Differentiating Composites of Standard Functions
Differentiating Composite Tangent Functions

III The Second Derivative

III.1 Second Derivative

Definition — Second Derivative

The second derivative of $f$, denoted $f''$, is the derivative of the first derivative, $f'$.$$ f''(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}(f'(x)) \quad \text{ or in Leibniz notation, } \quad \dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2}= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} \right) $$

Example

Find the second derivative of $f(x)=x^4 - 5x^2$.

Answer

First, we find the first derivative:$$ f'(x) = 4x^3 - 10x $$ Now, we differentiate again to find the second derivative:$$ f''(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}(4x^3 - 10x) = 12x^2 - 10 $$

Skills to practice

Calculating First and Second Derivatives
Showing That a Function Satisfies a Relation

IV Going further

Skills to practice

Solving Problems
Quiz