Chain Rule Questions

The Chain Rule is a fundamental concept in calculus that allows us to find the derivative of composite functions. By breaking down complex functions into smaller, more manageable parts.

1.0Differentiation using the Chain Rule

The chain rule of differentiation is a fundamental concept in calculus that allows us to find the derivative of a composite function. 

Here's the basic idea: Suppose you have a function f and another function g inside f, like f(g(x)). The chain rule states that to determine the derivative of f(g(x)) with respect to x, you need to first find the derivative of f with respect to g, and then multiply it by the derivative of g with respect to x.

Mathematically, If y = f(u) and u = g(x) then the chain rule is expressed as:

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

=f'(u)⋅g'(x) = f'(g(x))⋅g'(x). 

It can be extended to any number of chains.

2.0Chain Rule Differentiation Solved Examples

Example 1: y = tan(log x)

Solution:   

Given function is:

y = tan (log x)

Let y = tan t and t = log x

$\frac{dy}{dt}={\sec }^{2}t$

$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

Now, using the chain rule, we have.

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

$={\sec }^2(\log x)\cdot \frac{1}{x}$

Thus, the derivative of $y=tan(logx)is\frac{dy}{dx}={\sec }^2(\log x)\cdot \frac{1}{x}$

Example 2: y = sinx²

Solution: Given function is y = sinx²

Let y = sin t and t = x²

$\frac{\mathrm{dy}}{\mathrm{dt}}=\cos t,\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}$ .

Now, using the chain rule, we have.

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

$\frac{dy}{dx}=\cos \left(x^2\right)\cdot 2x$

= cosx².2x

Thus, the derivative of y = sinx² is $\frac{dy}{dx}={\cos }^2.2x$

Example 3: y = log sinx²

Solution: Given function is :  y = log sinx²

Let y = log t, t = sinx²

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{\mathrm{t}},\frac{\mathrm{dt}}{\mathrm{dx}}=\cos {\mathrm{x}}^2\cdot 2\mathrm{x}$

Now, using the chain rule, we have.

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}}\cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{dy}{dx}=\frac{1}{\sin x^2}\cdot \cos x^2\cdot 2x$

Thus, the derivative of y = log sinx² is $\frac{dy}{dx}=\frac{1}{\sin x^2}\cdot \cos x^2\cdot 2x$

Example 4: y = tan ^–1(log sinx²)]

Solution: Given function is:

y = tan^–1(log sinx²)]

Let y = tan^–1t, t = log sinx²

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{1+{\mathrm{t}}^2},\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{\sin {\mathrm{x}}^2}\cdot \cos {\mathrm{x}}^2\cdot 2\mathrm{x}$

Now, using the chain rule, we have;

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

$\frac{dy}{dx}=\frac{1}{1+\left(\log \sin x^2\right)}\cdot \frac{1}{\sin x^2}\cdot \cos x^2\cdot 2x$

Thus, the derivative of y = tan ^–1(log sinx²)] is $\frac{dy}{dx}=\frac{1}{1+\left(\log \sin x^2\right)}\cdot \frac{1}{\sin x^2}\cdot \cos x^2\cdot 2x$ .

Example 5: Find the derivative of y = e^{x sin x}

Solution:

Given function is:

y = e^{x sin x}

Let y = e^t and t = x sin x

$\frac{dy}{dt}=e^t$

$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}\sin \mathrm{x})$

= x cos x + sin x {from the product rule of differentiation}

Now, using the chain rule, we have;

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

= e^t (x cos x + sin x)

= e^{x sin x} (x cos x + sin x)

Thus, the derivative of y = e^{x sin x} is $\frac{\mathrm{dy}}{\mathrm{dx}}=exsinx(xcosx+sinx)$.

3.0Chain Rule Derivative Practice Questions

Differentiate

1. y = (2x³ + 5x)⁴
2. y = sin(3x² + x)
3. y = tan(2x + 3)
4. y = e^x log sin2x

Answer Sheet:

1. $\frac{dy}{dx}=4{\left(2x^3+5x\right)}^3\times \left(6x^2+5\right)$
2. $\frac{dy}{dx}=\cos \left(3x^2+1\right)\times (6x+1)$
3. $\frac{dy}{dx}={\sec }^2(2x+3)\times 2$
4. $\frac{dy}{dx}=e^x\times \log (\sin 2x)+\frac{e^x}{\sin 2x}\times 2\cos 2x$

4.0Solved Problems

1. What is the chain rule in differentiation?

Ans: The chain rule is a formula used to determine the derivative or differentiation of a composite function. Given two functions, f and g, and a composite function h(x) = f(g(x)), the chain rule asserts that the derivative of h(x) with respect to x is: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$

2. When should I use the chain rule?

Ans: Use the chain rule when you need to differentiate a function made up of two or more functions or we can say a function is a composite function. This typically happens when you have a function inside another function, like $\sin \left(x^2\right)\text{ or }{\mathrm{e}}^{\ln (\mathrm{x})}.$

3. How do I identify the inner and outer functions?

Ans: The inner function is the function inside another function, and the outer function is the function that is applied to the inner function. For example, in h(x) = sin(x²):

• Inner function: g(x) = x²
• Outer function: f(u) = sin(u), where u = g(x)

4. What is the chain rule for multiple variables?

Ans: For functions of multiple variables, the chain rule can be extended. If z = f (x, y), where x = g(t) and y = h(t), the chain rule states:

$\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\cdot \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\cdot \frac{\mathrm{dy}}{\mathrm{dt}}$