Question

If $u=f{\left(x\right)}^3,v=g\left(x^2\right),f\text{'}\left(x\right)=\cos x,g\text{'}\left(x\right)=\sin x$then the value of $\frac{du}{dv}$ is

• A. $\tan x$
• B. $\frac{3}{2}x\cos \left(x^3\right)\cos ec\left(x^2\right)$
• C. $\frac{2}{3}\sin x^3\sin x^2$
• D. None of these

Answer 1

The correct option is B

$\frac{3}{2}x\cos \left(x^3\right)\cos ec\left(x^2\right)$

Explanation for the correct answer:

To find the value of $\frac{du}{dv}$ :

Given,

$$\begin{gathered} u=f{\left(x\right)}^3,v=g\left(x^2\right),f\text{'}\left(x\right)=\cos x,g\text{'}\left(x\right)=\sin x \\ \frac{du}{dv}=\raisebox{1ex}{${\displaystyle \frac{du}{dx}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{dv}{dx}}$}\right. \\ f\text{'}\left(x^3\right)=\cos x^{3}andg\text{'}\left(x^2\right)=\sin x^2 \end{gathered}$$

Now, differentiate $u$ and $v$ with respect to $x$.

$\begin{array}{rcl}\frac{du}{dx}& =& f’\left(x^3\right)3x^2\\ \frac{du}{dx}& =& 3x^2.\cos x^3...\left(1\right)\\ & & \\ \frac{dv}{dx}& =& g\text{'}\left(x^2\right)2x\\ \frac{dv}{dx}& =& 2x\sin x^2...\left(2\right)\end{array}$

Divide $\left(1\right)$ by $\left(2\right)$

$\begin{array}{rcl}\frac{du}{dv}& =& \frac{[3x^2\cos x^3]}{[2x\sin x^2]}\\ & =& \left(\frac{3}{2}\right)\left(\frac{x\cos x^3}{\sin x^2}\right)\end{array}$

Hence, the correct option is B.