Chain Rule Examples

Example 1:

Differentiate y = {\left( {2{x^3} - 5{x^2} + 4} \right)^5} with respect to x using the chain rule method.

 

Let us consider u = 2{x^3} - 5{x^2} + 4, then y = {u^5}. Applying the chain rule, we have

\begin{gathered}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \times \frac{{du}}{{dx}} \\ \frac{{dy}}{{dx}} = 5{u^{5 - 1}} \times \frac{d}{{dx}}\left( {2{x^3} - 5{x^2} + 4} \right) \\ \frac{{dy}}{{dx}} = 5{u^4}\left( {6{x^2} - 10x} \right) \\ \frac{{dy}}{{dx}} = 5{\left( {2{x^3} - 5{x^2} + 4} \right)^4}\left( {6{x^2} - 10x} \right) \\ \end{gathered}

Example 2:

Differentiate y = {x^2} + 4 with respect to \sqrt {{x^2} + 1} using the chain rule method.

If u = \sqrt {{x^2} + 1} , then we have to find \frac{{dy}}{{du}}. Using the chain rule method

\frac{{dy}}{{du}} = \frac{{dy}}{{dx}} \times \frac{{dx}}{{du}}

First we differentiate the function y = {x^2} + 4 with respect to x.

\frac{{dy}}{{dx}} = 2x

Now differentiate the function u = \sqrt {{x^2} + 1} with respect to x.

\frac{{du}}{{dx}} = \frac{x}{{\sqrt {{x^2} + 1} }}

Now using the chain rule of differentiation, we have

\begin{gathered}\frac{{dy}}{{du}} = \frac{{dy}}{{dx}} \times \frac{{dx}}{{du}} \\ \frac{{dy}}{{du}} = 2x \times \frac{{\sqrt {{x^2} + 1} }}{x} \\ \frac{{dy}}{{du}} = 2\sqrt {{x^2} + 1} \\ \end{gathered}