Question

If $u=f\left(x^3\right),v=g\left(x^2\right),f^{\prime }\left(x\right)=cosx$ and $g^{\prime }\left(x\right)=sinx$, then find $\frac{du}{dv}$

• A. $\frac{3}{2}{x}^2\cdot cosec{x}^3\cdot \cos x^2$
• B. $\frac{2}{3}x\cdot cosec{x}^3\cdot \cos x^2$
• C. $\frac{4}{5}{x}^2\cdot \cos x^3\cdot cosec{x}^2$
• D. $\frac{3}{2}x\cdot \cos x^3\cdot cosec{x}^2$

Answer 1

Answer: (D)

$\frac{3}{2}x\cdot \cos x^3\cdot cosec{x}^2$

Here, $u=f\left(x^3\right)$
$\Rightarrow \frac{du}{dx}=f^{\prime }\left(x^3\right)\cdot \frac{d}{dx}\left(x^3\right)=\left\{cos\right(x^3\left)\right\}\cdot 3x^2=3x^2\cdot cos{x}^3$

and $v=g\left(x^2\right)$

$\Rightarrow \frac{dv}{dx}=g^{\prime }\left(x^2\right)\cdot \frac{d}{dx}\left(x^2\right)=\left\{sin{x}^2\right\}\cdot \left(2x\right)=2x\cdot sin{x}^2$

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{3x^2\cdot cos{x}^3}{2x\cdot sin{x}^2}\Rightarrow \frac{du}{dv}=\frac{3}{2}x\cdot cos{x}^3\cdot cosec{x}^2$