Question

If $x={\int }_{c^2}^{\tan t}{\tan }^{-1}zdz,y={\int }_n^{\sqrt{t}}\frac{\cos \left(z^2\right)}{z}dz$ than $\frac{dy}{dx}$ is equal to

• A. $\frac{\tan t}{2t}$
• B. $\frac{{\cos }^{2}t}{t^2}$
• C. $\frac{{\cos }^{3}t}{2t^2}$
• D. $\frac{\tan t^2}{2t^2}$

Answer 1

Answer: (C)

$\frac{{\cos }^{3}t}{2t^2}$

$x={\int }_{c^2}^{\tan t}{\tan }^{-1}zdz\Rightarrow \frac{dx}{dt}=t{\sec }^{2}t$
$y={\int }_n^{\sqrt{t}}\frac{\cos \left(z^2\right)}{z}dz\Rightarrow \frac{dy}{dt}=\frac{1}{2}\left(\frac{\cos t}{\sqrt{t}}\right)\frac{1}{\sqrt{t}}$
$\therefore \frac{dy}{dx}=\frac{{\cos }^{3}t}{2t^2}$