Question

If $u=f(x^2),v=g(x^3),f^{\prime }(x)=\sin x$ and $g^{\prime }(x)=\cos x,$ 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. $\frac{3x\sin x^2}{2cos{x}^3}$

Answer 1

Answer: (B)

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