Let \[ f(x) = (g_{\circ} h)(x) \\ = g(h(x) \]
be the composition of two functions.
Let \( u = h(x) \). Using the above, function \( f \) may be written as:
\[ f(x) = g(u)\]
the derivative \( f' \) of \( f \) with respect to \( x \) is given by the chain rule of differentiation [1] :
\[ \boxed { f'(x) = \dfrac{df}{du} \dfrac{du}{dx} } \]

Examples Using the Chain Rule of Differentiation

We now present several examples of applications of the chain rule.

Example 1

Find the derivative \( f'(x) \) given
\[ f(x) = 4 \cos (5x - 2) \]

Solution to Example 1

Let \( u = 5x - 2 \) and \( f(u) = 4 \cos u \), hence
\[ \dfrac{du}{dx} = 5 \] and \[ \dfrac{df}{du} = - 4 \sin u \]
We now use the chain rule
\[ f '(x) = \dfrac{df}{du} \dfrac{du}{dx} \\ = - 4 \sin (u) \cdot 5\]
We now substitute \( u = 5x - 2\) in \( \sin (u) \) above to obtain
\[ \boxed{ f '(x) = - 20 \sin (5x - 2) } \]

Example 2

Find the derivative \( f '(x) \) given
\[ f(x) = (x^3 - 4x + 5)^4 \]

Solution to Example 2

Let \( u = x^3 - 4x + 5\) and \( f(u) = u^4 \) which give
\[ \dfrac{du}{dx} = 3 x^2 - 4 \] and \[ \dfrac{df}{du} = 4 u^3 \]
Use of the chain rule
\[ f '(x) = \dfrac{df}{du} \dfrac{du}{dx} \\
= (4 u^3) (3 x^2 - 4) \]
We now substitute \( u = x^3 - 4x + 5 \) above to obtain
\[ \boxed{ f '(x) = 4 (x^3 - 4x + 5)^3 (3 x^2 - 4) } \]

Example 3

Find \( f '(x) \) given
\[ f(x) = \sqrt {x^2 + 2x -1} \]

Solution to Example 3

Let \( u = x^2 + 2x -1 \) and \( f(u) = \sqrt u \) which give
\[ \dfrac{du}{dx} = 2x + 2 \] and \[ \dfrac{df}{du} = \dfrac{1}{2 \sqrt u} \]
Use the chain rule we obtain
\[ f '(x) = \dfrac{df}{du} \dfrac{du}{dx} = \dfrac{1}{ 2 \sqrt u} (2x+2) \]
Substitute \( u = x^2 + 2x -1 \) above to obtain
\[ f '(x) = \dfrac{2x + 2}{2 \sqrt{x^2 + 2x -1}} \]
Factor 2 in numerator and denominator and simplify
\[ \boxed{ f '(x) = \dfrac{x + 1}{ \sqrt{x^2 + 2x -1}}} \]

Example 4

Find the first derivative of \( f \) given
\[ f(x) = \sin ^2 (2x + 3) \]

Solution to Example 4

Let \( u = \sin (2x + 3) \) and \( f(u) = u^2 \) which give
\[ \dfrac{du}{dx} = 2 \cos(2x + 3) \] and \[ \dfrac{df}{du} = 2 u \]
The use of the chain rule leads to
\[ f '(x) = \dfrac{df}{du} \dfrac{du}{dx} = 2 u \cdot 2 \cos(2x + 3) \]
Substitute \( u = \sin (2x + 3) \) above to obtain
\[ f '(x) = 4 \sin (2x + 3) \cos (2x + 3) \]
Use the trigonometric identity \( \sin (2x) = 2 \sin x \cos x \) to simplify \( f '(x) \)
\[ \boxed{ f '(x) = 2 \sin (4x + 6) } \]

Example 5

Find the first derivative of \( f \) given
\[ f(x) = \ln(x^2 + x) \]

Solution to Example 5

Let \( u = x^2 + x \) and \( f(u) = \ln u \) , hence
\[ \dfrac{du}{dx} = 2 x + 1 \] and \[ \dfrac{df}{du} = \dfrac{1}{u} \]
Use the chain rule and substitute
\[ f '(x) = \dfrac{df}{du} \dfrac{du}{dx} = \dfrac{1}{u} (2 x + 1) \]
Substitute back \( u = x^2 + x \)
\[ \boxed{ f'(x) = \dfrac{2 x + 1}{x^2 + x} } \]

Exercises On Chain Rule of Differentiation

Use the chain rule to find the first derivative to each of the functions.