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