Chain Rule for Differentiation"

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 Chain Rule
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Chain Rule:
If $y = f\big(g(x)\big)$, where both $f$ and $g$ are differentiable functions, then the derivative of $y$ with respect to $x$ is given by:

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

This means that when differentiating a composite function, we differentiate the outer function first, leaving the inner function unchanged, and then multiply by the derivative of the inner function.

Example 1: Differentiating a Square Root Function

Differentiate $f(x) = \sqrt{4x^2 + 1}$.

Solution:

Rewrite as:

$f(x) = (4x^2 + 1)^{\frac{1}{2}}$.

Using the chain rule, let:

• Outer function: $f(u) = u^{\frac{1}{2}}$, so $f'(u) = \frac{1}{2} u^{-\frac{1}{2}}$.
• Inner function: $g(x) = 4x^2 + 1$, so $g'(x) = 8x$.

Applying the chain rule:

$f'(x) = \dfrac{1}{2} (4x^2 + 1)^{-\frac{1}{2}} \cdot 8x$

$f'(x) = \dfrac{8x}{2\sqrt{4x^2 + 1}}$

$f'(x) = \boxed{\dfrac{4x}{\sqrt{4x^2 + 1}}}$.

Example 2: Differentiating an Exponential Function

Differentiate $f(x) = e^{3x^2 + 2x}$.

Solution:

Using the chain rule, let:

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

Applying the chain rule:

$f'(x) = e^{3x^2 + 2x} \cdot (6x + 2)$

Thus, the final answer is:

$\boxed{(6x + 2).e^{3x^2 + 2x}}$.

Example 3: Differentiating a Trigonometric Function

Differentiate $f(x) = \sin(5x^3 - 2x)$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = \sin u$, so $f'(u) = \cos u$.
• Inner function: $g(x) = 5x^3 - 2x$, so $g'(x) = 15x^2 - 2$.

Applying the chain rule:

$f'(x) = \cos(5x^3 - 2x) \cdot (15x^2 - 2)$

Thus, the final answer is:

$\boxed{(15x^2 - 2) .\cos(5x^3 - 2x)}$.

Example 4: Finding the Rate of Change of Temperature

The temperature $T$ of an object is given by: $T = \ln(5t^2 + 3t)$

where $t$ is the time in seconds. Find the rate of change of temperature at $t = 2$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = \ln u$, so $f'(u) = \dfrac{1}{u}$.
• Inner function: $g(t) = 5t^2 + 3t$, so $g'(t) = 10t + 3$.

Applying the chain rule:

$\dfrac{dT}{dt} = \frac{1}{5t^2 + 3t} \cdot (10t + 3)$

Substituting $t = 2$:

$\dfrac{dT}{dt} = \dfrac{10(2) + 3}{5(2)^2 + 3(2)} = \dfrac{20 + 3}{20 + 6} = \dfrac{23}{26}$.

Thus, the rate of change of temperature at $t = 2$ is: $\boxed{\frac{23}{26}}$.