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