Chain Rule Examples

Example 1:
Differentiate y =  {\left( {2{x^3} - 5{x^2} + 4} \right)^5} with respect to x using chain rule method.

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

If u = \sqrt {{x^2} + 1} , then we have to find \frac{{dy}}{{du}}. By 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}

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