Chain Rule And Composite Functions

Derivative of Composite Function with the help of chain rule:

When two functions are combined in such a way that the output of one function becomes the input to another function then this is referred to as composite function.

A composite function is denoted as:

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

For finding the derivative of a composite function \(f(g(x))\) where both the functions \(f(x)\) and \(g(x)\) are differentiable then the derivative of the composite function is given as,

\( (fog)’ = (f’og) × g'\)

To illustrate this concept let us go through an example. Suppose we have to find the derivative of where \(f(x)\) = \( e^{x^2 + 4}\).

Now using the concept of composite function,

Given \(f(x)\) = \(e^{x^2 + 4}\)

Let \(g(x)\) = \(x^2 + 4 \)

Since \( (fog)’ \) = \( (f’og) × g’ \)

Let  \(g(x)\) =\( k\) then \(f(x)\) = \(e^k\)

\( \Rightarrow (f’og)\) = \(e^k\) and \( g'\) = \( 2x \)

\(\Rightarrow (fog)'\) = \(e^k × 2x \)

\(\Rightarrow \frac{d(f(x))}{dx}\) =\( e^{x^2 + 4} × 2x \)

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:

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

Now if the functions u and v are differentiable and \(\frac{dt}{dx}\) and \(\frac{dv}{dt}\) exists, then the composite function \(f(x)\) is also differentiable and is given as following,

Let \( t = u(x) \) and if there exists \( \frac{dt}{dx} \) and \(\frac{dv}{dt}\) then using Leibnitz notation this can be expressed as

\( \frac{df}{dx} \) = \( \frac{dv}{dt} × \frac{dt}{dx} \)

Let us illustrate it with the help of an example:

Example 1: Find the derivative of the function \(f(x)\) = \(sin(2x^2 – 6x)\).

Solution 1: The given function represents a composite function where

\( f(x) \) =\( sin(2x^2 – 6x)\)

\( u(x) \) =\( 2x^2 – 6x\)

\( v(t) \) = \(sin t\)

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

According to the chain rule,

Chain RuleIn layman terms to differentiate a composite function at any point in its domain first differentiate the outer function (i.e. the function enclosing some other function) and then multiply it with the derivative of the inner function to get the desired differentiation.

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

Example 2: Find the derivative of the function given by \(f(x)\) = \(sin(e^{x^3})\)

Solution 2: It is a composition of three functions namely \(p(s)\) = \(sin ~s\), \(q(t)\) =\( e^t\) and \(r(x)\) = \(x^3\).

\(\Rightarrow f(x)\) = \(p(q(r(x))) \)

By chain rule we can solve the above function as,

\(\frac{df}{dx}\) = \(\frac{dp}{ds} × \frac{ds}{dt} × \frac{dt}{dx}\) = \(cos~ s × e^t × 3x^2\)

= \(cos e^{x^3} × e^{x^3} × 3x^2\)

Example 3: Find the derivative of the function given by

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

Solution 3: 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 \(t\) = \(u(x)\)

\(\Rightarrow \frac{dt}{dx} \)= \(2x\)

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

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

To practice more on chain rule and differentiation of composite functions,  download BYJU’s-the learning app to excel in knowledge.


Practise This Question

If y=[x+x2+a2]n,then dydx is equal to