Differentiation of Composite Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Derivative of Composite Functions
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Derivative of Composite Functions:
If a function $y = f\big(g(x)\big)$ is a composite function, where $g(x)$ is the inner function and $f(x)$ is the outer function, then the derivative of $y$ with respect to $x$ is given by:

$\boxed{\frac{dy}{dx} = f'\big(g(x)\big) \cdot g'(x)}$

This is known as the chain rule and is used to differentiate composite functions, where we differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function.

Example 1: Differentiating a Polynomial Inside a Power Function

Differentiate $f(x) = (3x^2 + 2x)^4$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = u^4$, so $f'(u) = 4u^3$.
• Inner function: $g(x) = 3x^2 + 2x$, so $g'(x) = 6x + 2$.

Applying the chain rule:

$f'(x) = 4(3x^2 + 2x)^3 \cdot (6x + 2)$

Thus, the final answer is: $\boxed{4.(3x^2 + 2x)^3 .(6x + 2)}$.

Example 2: Differentiating a Trigonometric Function Inside an Exponential

Differentiate $f(x) = e^{\sin(x)}$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = e^u$, so $f'(u) = e^u$.
• Inner function: $g(x) = \sin(x)$, so $g'(x) = \cos(x)$.

Applying the chain rule:

$f'(x) = e^{\sin(x)} \cdot \cos(x)$

Thus, the final answer is: $\boxed{e^{\sin(x)} .\cos(x)}$.

Example 3: Differentiating a Logarithmic Function Inside a Polynomial

Differentiate $f(x) = \ln(5x^3 + 3x + 1)$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = \ln(u)$, so $f'(u) = \dfrac{1}{u}$.
• Inner function: $g(x) = 5x^3 + 3x + 1$, so $g'(x) = 15x^2 + 3$.

Applying the chain rule:

$f'(x) = \dfrac{1}{5x^3 + 3x + 1} \cdot (15x^2 + 3)$

Thus, the final answer is: $\boxed{\frac{15x^2 + 3}{5x^3 + 3x + 1}}$.

Example 4: Finding the Rate of Change of Velocity in Physics

The velocity $v(t)$ of a particle is given by: $v(t) = (3t^2 + 1)^{5}$.

Find the rate of change of velocity (i.e., acceleration) at time $t = 2$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = u^5$, so $f'(u) = 5u^4$.
• Inner function: $g(t) = 3t^2 + 1$, so $g'(t) = 6t$.

Applying the chain rule:

$a(t) = \dfrac{d}{dt} (3t^2 + 1)^5 = 5(3t^2 + 1)^4 \cdot (6t)$

Substitute $t = 2$:

$a(2) = 5(3(2)^2 + 1)^4 \cdot (6(2))$

$a(2) = 5(12 + 1)^4 \cdot 12 = 5(13)^4 \cdot 12 = 1713660$

Thus, the acceleration at $t = 2$ is: $\boxed{1713660}$.