Chain Rule of Differentiation in Calculus

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:

\[ \dfrac{du}{dx} = 5 \quad \text{and} \quad \dfrac{df}{du} = -4 \sin u \]

We now use the chain rule:

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = -4 \sin(u) \cdot 5 \]

Substitute \( u = 5x - 2 \) back into the expression:

\[ f'(x) = -20 \sin (5x - 2) \]

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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:

\[ \dfrac{du}{dx} = 3x^2 - 4 \quad \text{and} \quad \dfrac{df}{du} = 4u^3 \]

Use of the chain rule:

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = (4u^3)(3x^2 - 4) \]

Substitute \( u = x^3 - 4x + 5 \) back:

\[ f'(x) = 4(x^3 - 4x + 5)^3 (3x^2 - 4) \]

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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:

\[ \dfrac{du}{dx} = 2x + 2 \quad \text{and} \quad \dfrac{df}{du} = \dfrac{1}{2\sqrt{u}} \]

Use the chain rule:

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = \dfrac{1}{2\sqrt{u}} (2x + 2) = \dfrac{2x + 2}{2\sqrt{x^2 + 2x - 1}} \]

Factor 2 in numerator and denominator and simplify:

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

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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:

\[ \dfrac{du}{dx} = 2 \cos(2x + 3) \quad \text{and} \quad \dfrac{df}{du} = 2u \]

The use of the chain rule leads to:

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = 2u \cdot 2 \cos(2x + 3) = 4 \sin (2x + 3) \cos (2x + 3) \]

Use the trigonometric identity \( \sin(2x) = 2 \sin x \cos x \) to simplify:

\[ f'(x) = 2 \sin (4x + 6) \]

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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:

\[ \dfrac{du}{dx} = 2x + 1 \quad \text{and} \quad \dfrac{df}{du} = \dfrac{1}{u} \]

Use the chain rule and substitute back \( u = x^2 + x \):

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = \dfrac{1}{u} (2x + 1) = \dfrac{2x + 1}{x^2 + x} \]

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

1. \( f(x) = \cos (3x - 3) \)
2. \( l(x) = (3x^2 - 3x + 8)^4 \)
3. \( m(x) = \sin \left(\dfrac{1}{x-2}\right) \)
4. \( t(x) = \sqrt{3x^2 - 3x + 6} \)
5. \( r(x) = \sin^2 (4x + 20) \)

Answers to the Above Exercises

1. \( f'(x) = -3 \sin (3x - 3) \)
2. \( l'(x) = 12(2x - 1)(3x^2 - 3x + 8)^3 \) (Note: factoring gives \(12(2x-1)\) from \(4(3x^2-3x+8)^3(6x-3)\))
3. \( m'(x) = -\dfrac{1}{(x - 2)^2} \cos \left(\dfrac{1}{x-2}\right) \)
4. \( t'(x) = \dfrac{6x - 3}{2\sqrt{3x^2 - 3x + 6}} \)
5. \( r'(x) = 8\sin(4x+20)\cos(4x+20) = 4\sin(8x+40) \)