The Chain Rule

Introduction

1.Let $g$ be differentiable at $x$, and let $f$ be differentiable at $z = g(x)$. Then the composite function $y = f(g(x))$ is differentiable at $x$, and its derivative is the product of the derivative of $f$ evaluated at $g(x)$ and the derivative of $g$ at $x$:

$$D[f(g(x))] = f'(g(x)) \cdot g'(x)$$

2.This result is known as the chain rule. It states that to differentiate a composite function, one multiplies the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.

3.In Leibniz notation, if $y = f(u)$ and $u = g(x)$, the chain rule takes the form:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Proof

4.Fix the point $x$ and write $z = g(x)$. Differentiating the composition directly from the limit is delicate, because the natural manipulation of the difference quotient divides by $g(x+h) - g(x)$, a quantity that can vanish for values of $h$ arbitrarily close to $0$. To sidestep this, we record the derivative of $f$ at $z$ in an auxiliary function that involves no division. Define $\varphi$ on the domain of $f$ by:

$$ \varphi(w) = \begin{cases} \dfrac{f(w) - f(z)}{w - z} & w \neq z \\[8pt] f'(z) & w = z \end{cases} $$

5.Since $f$ is differentiable at $z$, the difference quotient tends to $f'(z)$ as $w \to z$, which is the value assigned to $\varphi$ at $z$. The auxiliary function is therefore continuous at $z$:

$$\lim_{w \to z} \varphi(w) = f'(z) = \varphi(z)$$

6.Directly from its definition, $\varphi$ satisfies the identity:

$$f(w) - f(z) = \varphi(w)(w - z)$$

7.For $w \neq z$ this is the definition of $\varphi$ multiplied by $w - z$, and for $w = z$ both sides vanish, so the identity holds for every $w$ without exception.

8.Evaluating the identity at $w = g(x+h)$, for which $w - z = g(x+h) - g(x)$, and dividing by $h \neq 0$, the difference quotient of the composite function splits into a product:

$$\frac{f(g(x+h)) - f(g(x))}{h} = \varphi(g(x+h)) \cdot \frac{g(x+h) - g(x)}{h}$$

9.Now let $h \to 0$. The inner function $g$ is differentiable at $x$, hence continuous there, so $g(x+h) \to g(x) = z$. Since $\varphi$ is continuous at $z$, the composition gives $\varphi(g(x+h)) \to \varphi(z) = f'(z)$. The second factor is the difference quotient of $g$ at $x$, which tends to $g'(x)$. Therefore:

$$ \begin{align} D[f(g(x))] &= \lim_{h \to 0} \left( \varphi(g(x+h)) \cdot \frac{g(x+h) - g(x)}{h} \right) \\[6pt] &= f'(z) \cdot g'(x) \\[6pt] &= f'(g(x)) \cdot g'(x) \end{align} $$

10.

The identity $f(w) - f(z) = \varphi(w)(w - z)$ is the step the naive computation cannot supply. It multiplies by $w - z$ instead of dividing, so it stays valid at the values of $h$ where $g(x+h) = g(x)$, which is exactly where dividing by $g(x+h) - g(x)$ would break down.

A first example

11.Let us compute the derivative of the following composite function:

$$y = f(g(x)) = \sin(3x^2 + 2x)$$

12.In this case, we have:

14.The derivative of the outer function $f(t) = \sin(t)$ is:

$$f'(t) = \cos(t)$$

15.Substituting $t = g(x)$, we obtain:

$$f'(g(x)) = \cos(3x^2 + 2x)$$

16.The derivative of the inner function $g(x) = 3x^2 + 2x$ is:

$$g'(x) = 6x + 2$$

17.Applying the chain rule, we multiply the two derivatives:

$$D[f(g(x))] = f'(g(x)) \cdot g'(x) = (6x + 2)\cos(3x^2 + 2x)$$

18.We conclude that the derivative of $y = \sin(3x^2 + 2x)$ is $(6x + 2)\cos(3x^2 + 2x)$.

The chain rule in Leibniz notation

19.In Leibniz notation the computation proceeds by an explicit change of variable. Consider the function:

$$y = \ln(x^2 + 1)$$

20.Setting $u = x^2 + 1$, the function decomposes as $y = \ln(u)$ with $u = x^2 + 1$. The two derivatives are:

$$\frac{dy}{du} = \frac{1}{u}, \qquad \frac{du}{dx} = 2x$$

21.By the chain rule, the derivative of $y$ with respect to $x$ is the product of the two:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x$$

22.Since the result must be expressed in terms of the original variable, we substitute $u = x^2 + 1$ back into the expression:

$$\frac{dy}{dx} = \frac{2x}{x^2 + 1}$$

23.

The substitution of $u = g(x)$ at the end of the computation corresponds, in the prime notation, to evaluating $f'$ at $g(x)$ rather than at $x$. Omitting this step is among the most frequent errors in applying the rule.

Special cases

24.When the outer function is a power function, an exponential, or a logarithm, the chain rule produces formulas that recur constantly in computations. For a differentiable function $f$:

$$ \begin{align} D[f(x)^a] &= a[f(x)]^{a-1}f'(x) \\[14pt] D\left[e^{f(x)}\right] &= e^{f(x)}f'(x) \\[6pt] D[\ln f(x)] &= \frac{f'(x)}{f(x)} \end{align} $$

25.The first formula holds for any real exponent $a$, and the third requires $f(x) > 0$. Each follows from the chain rule with outer function $t^a$, $e^t$ and $\ln(t)$ respectively.

26.For example, in order to differentiate $y = (x^3 + 2)^{50}$, there is no need to expand the power. Applying the first formula with $f(x) = x^3 + 2$ and $a = 50$, we obtain:

$$D[(x^3 + 2)^{50}] = 50(x^3 + 2)^{49} \cdot 3x^2 = 150x^2(x^3 + 2)^{49}$$

27.The power formula also applies to irrational functions expressed through roots, since $\sqrt{f(x)} = f(x)^{1/2}$. To differentiate $y = \sqrt{x^2 + 1}$, we apply the formula with $f(x) = x^2 + 1$ and $a = \frac{1}{2}$:

$$D\left[\sqrt{x^2 + 1}\right] = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$$

28.We conclude that the derivative of $y = \sqrt{x^2 + 1}$ is $\dfrac{x}{\sqrt{x^2 + 1}}$.

29.

The power formula applies only when the exponent is constant. A related technique handles composite power functions, that is, functions of the type $f(x)^{g(x)}$, where both the base and the exponent depend on $x$.

Combining with the product rule

30.In most computations the chain rule operates together with the product or quotient rule rather than alone. The basic differentiation rules handle the algebraic structure of an expression, while the chain rule handles the nesting of one function inside another. Consider the function:

$$y = x^2 e^{\sin(x)}$$

31.The function is a product of $x^2$ and $e^{\sin(x)}$, so the differentiation begins with the product rule:

$$D\left[x^2 e^{\sin(x)}\right] = 2xe^{\sin(x)} + x^2 D\left[e^{\sin(x)}\right]$$

32.The remaining factor $e^{\sin(x)}$ is a composite function with outer function $e^t$ and inner function $\sin(x)$, so the chain rule gives:

$$D\left[e^{\sin(x)}\right] = e^{\sin(x)}\cos(x)$$

33.Substituting this result into the previous expression and collecting the common factor $xe^{\sin(x)}$, we obtain:

$$D\left[x^2 e^{\sin(x)}\right] = 2xe^{\sin(x)} + x^2 e^{\sin(x)}\cos(x) = xe^{\sin(x)}(2 + x\cos(x))$$

34.We conclude that the derivative of $y = x^2 e^{\sin(x)}$ is $xe^{\sin(x)}(2 + x\cos(x))$.

Extension to multiple compositions

35.The chain rule can be extended to compositions involving three or more functions. For example, given $y = f(g(h(x)))$, the derivative is:

$$D[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$

36.Each factor is the derivative of a function in the composition, evaluated at the composition of all subsequent functions. This pattern generalises to any finite number of nested functions. For $y = f_1(f_2(\cdots f_n(x)\cdots))$, the derivative is given by the product:

$$f_1'(f_2(\cdots f_n(x)\cdots)) \cdot f_2'(f_3(\cdots f_n(x)\cdots)) \cdots f_{n-1}'(f_n(x)) \cdot f_n'(x)$$

37.In practical applications, differentiation proceeds from the outermost function inward, with each derivative computed in sequence and the results multiplied together.

38.As an example, consider $y = \sin(e^{3x})$. The composition involves three functions:

$$ \begin{align} h(x) &= 3x \\[6pt] g(t) &= e^t \\[6pt] f(s) &= \sin(s) \end{align} $$

39.Applying the chain rule from the outside inward we obtain:

$$ \begin{align} D[\sin(e^{3x})] &= \cos(e^{3x}) \cdot e^{3x} \cdot 3 \\[6pt] &= 3e^{3x}\cos(e^{3x}) \end{align} $$

40.We conclude that the derivative of $y = \sin(e^{3x})$ is $3e^{3x}\cos(e^{3x})$.