Question

If $u = f(x^2) , v = g(x^3) , f'(x) = \sin x $ and $g'(x) = \cos x,$ then $ \frac{du}{dv} = $

• A. $\frac{3x \cos x^{3}}{2 \sin x^{2}}$
• B. $\frac{2 \sin x^{2}}{3x \cos x^{3}}$
• C. $\frac{2 \sin x^{2}}{3 \cos x^{3}}$
• D. $\frac{3x \sin x^{2}}{{2} cosx^{3}}$

Answer 1

Answer: (B)

$u =f\left(x^2\right) , v =g\left(x^{3}\right) $
$f'\left(x\right) = \sin x, g'\left(x\right) =\cos x$
$ \frac{du}{dx} = f'\left(x^{2}\right).2x$ and $\frac{dv}{dx} =g'\left(x^{3}\right) .3x^{2}$
$ \frac{du}{dv} =\frac{du}{dx}. \frac{dx}{dv} = \frac{f'\left(x^{2}\right).2x}{g'\left(x^{3}\right).3x^{2}} = \frac{2}{3x} \frac{ \sin x^{2}}{\cos x^{3}} $