Question

If $y=\left(\frac{1}{4}\right)u^4$, $u=\frac{2}{3}{x}^3+5$, then $\frac{dy}{dx}=$

• A. $\frac{x^2}{27}{\left(2x^3+15\right)}^3$
• B. $\frac{2x}{27}{\left(2x^3+5\right)}^3$
• C. $\frac{2x^2}{27}{\left(2x^3+15\right)}^3$
• D. None of these

Answer 1

The correct option is C

$\frac{2x^2}{27}{\left(2x^3+15\right)}^3$

Explanation for the correct option.

Step 1: Differentiate $y$ with respect to $u$

$\begin{array}{rcl}y& =& \left(\frac{1}{4}\right)u^4\\ & \Rightarrow & \frac{dy}{du}=\frac{1}{4}\left(4u^3\right)\\ & =& u^3\end{array}$

Step 2: Differentiate $u$ with respect to $x$

$\begin{array}{rcl}u& =& \frac{2}{3}{x}^3+5\\ & \Rightarrow & \frac{du}{dx}=\frac{2}{3}\left(3x^2\right)+0\\ & =& 2x^2\end{array}$

Step 3: Find $\frac{dy}{dx}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{du}\times \frac{du}{dx}\\ & =& u^3\times 2x^2\\ & =& {\left(\frac{2}{3}{x}^3+5\right)}^3\times 2x^2\\ & =& {\left(\frac{2x^3+15}{3}\right)}^3\times 2x^2\\ & =& \frac{1}{27}{\left(2x^3+15\right)}^3\times 2x^2\\ & =& \frac{2x^2}{27}{\left(2x^3+15\right)}^3\end{array}$

Hence, option C is correct.