 # Chain Rule And Composite Functions

In mathematics, we may come across different formulas and rules. Similarly, in differential calculus, chain rule is a formula used to find the derivative of a composite function. So before starting the formula of chain rule, let us understand the meaning of composite function and how it can be differentiated.

## What is Composite Function?

A composite function is denoted as:

(fog)(x) = f(g(x))

Suppose f(x) and g(x) are two differentiable functions such that the derivative of a composite function f(g(x)) can be expressed as

(fog)′ = (f′og) × g′

This can be understood in a better way from the example given below:

Consider f(x) = ex2 + 4 and g(x) = x2 + 4

Therefore, f'(x) = 2x ex2 and g'(x) = 2x

Now, the derivative of composite function of f(x) and g(x) can be written as:

(fog)′  = (f′og) × g′

Let  g(x) = k then f(x) = ek {where k = x2 + 4}

⇒ (f′og)  = ek and g′ = 2x

⇒(fog)′  = ek × 2x = ex2 + 4  × 2x

## Chain Rule

The rule applied for finding the derivative of composition of function is basically known as the chain rule. Let f represent a real valued function which is a composition of two functions u and v such that:

f = v(u(x))

Let us assume u(x) = t

Now if the functions u and v are differentiable and dt/dx and dv/dt exist, then the composite function f(x) is also differentiable. This can be done as given below.

Using Leibnitz notation, we can express the differentiation of the above function as

df/dx = (dv/dt) × (dt/dx)

In layman terms, to differentiate a composite function at any point in its domain, first differentiate the outer part (i.e. the function enclosing some other function) and then multiply it with the inner function’s derivative function. This will provide us with the desired differentiation.

As the name suggests, chain rule means differentiating the terms one by one in a chain form, starting from the outermost function to the innermost function.

Let us illustrate this rule with the help of an example:

Example 1:

Find the derivative of the function f(x) = sin(2x2 – 6x).

Solution:

The given can be expressed as a composite function as given below:

f(x) = sin(2x2 – 6x)

u(x) =2x2 – 6x

v(t) = sin t

Thus, t = u(x) = 2x2 – 6x

⇒f(x) = v(u(x))

According to the chain rule,

df/dx = (dv/dt) × (dt/dx)

Where,

dv/dt = d/dt (sin t) = cos t

dt/dx = d/dx [u(x)] = d/dx (2x2 – 6x) = 4x – 6

Therefore, df/dx = cos t × (4x – 6)

= cos(2x2 – 6x) × (4x – 6)

= (4x – 6) cos(2x2 – 6x)

## Chain Rule Formula

The formula of chain rule for the function y = f(x), where f(x) is a composite function such that x = g(t), is given as: This is the standard form of chain rule of differentiation formula.

Let us illustrate it with the help of an example:

### Chain Rule Examples

Question 1:

Find the derivative of the function given by $$f(x)$$ = $$sin(e^{x^3})$$

Solution:

Given,

$$f(x)$$ = $$sin(e^{x^3})$$

It is a composition of three functions such as:

p(s) = sin s, q(t) = et and r(x) = x3

Thus, f(x) = p(q(r(x)))

That means, t = x3 and s = ex3

Using chain rule formula,

df/dx = (dp/ds) × (ds/dt) × (dt/dx)

= [d/ds (sin s)] × [d/dt (et)] × [d/dx (x3)]

= cos s × et × 3x2

= $$cos e^{x^3} × e^{x^3} × 3x^2$$

Question 2:

Find the derivative of the function given by $$f(x)$$ = $$\sqrt{tan (x^2 + 1)}$$

Solution:

Given,

$$f(x)$$ = $$\sqrt{tan (x^2 + 1)}$$

The given function represents a composition of functions where

$$f(x)$$ = $$\sqrt{tan (x^2 + 1)}$$

$$u(x)$$= $$x^2 + 1$$

$$v(t)$$ =$$\sqrt{tan t}$$

$$\Rightarrow f(x)$$ =$$v(u(x))$$

According to the chain rule,

$$\frac{df}{dx}$$ = $$\frac{dv}{dt}$$ × $$\frac{dt}{dx}$$

$$\Rightarrow \frac{dv}{dt}$$ =$$\frac{1}{2} (tan t)^{\frac{-1}{2}}× sec^2 t$$
=$$\frac{1}{2} × \frac{1}{\sqrt{tan t}} × sec^2 t$$

Also $$t$$ = $$u(x)$$

$$\Rightarrow \frac{dt}{dx}$$= $$2x$$

$$\Rightarrow \frac{df}{dx}$$ =$$2x × \frac{1}{2} × \frac{1}{\sqrt{tan t}} × sec^2 t$$

Therefore,

$$\frac{df}{dx}$$ = $$\frac{x sec^2(x^2 + 1)}{\sqrt{tan(x^2 + 1)}}$$

### Chain Rule for Partial Derivatives

The chain rule for total derivatives implies a chain rule for partial derivatives. We know that the partial derivative in the ith coordinate direction can be evaluated by multiplying the ith basis vector’s Jacobian matrix when the total derivative exists. Hence, the chain rule for the function y = f(u) = (f1(u), …, fk(u)) and u = g(x) = (g1(x), …, gm(x)) can be written for partial derivatives as: ### Chain Rule Problems

Practice the question given below:

1. Find the derivative of the function y = cos2(x4)
2. Using chain rule, find the derivative of y = sin4x + sin x4
3. Find the derivative of y = 2 ln[ln(ln sec x)]

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