Question

If $x=\sin h^{-1}[\log (1+\sqrt{y})]$, then $\frac{dy}{dx}=$

• A. $2(y+\sqrt{y})\sin hx$
• B. $2(y+\sqrt{y})\sqrt{1-(\log (1+\sqrt{y}){)}^2}$
• C. $2(y+\sqrt{y})\cos hx$
• D. $2(y+\sqrt{y})\log (1+\sqrt{y})$

Answer 1

Answer: (A)

Given,
$x=\sin h^{-1}[\log (1+\sqrt{y})]$
$\sin hx=\log (1+\sqrt{y})$
$\cos hx=\frac{1}{(1+\sqrt{y})}\cdot \frac{1}{2\sqrt{y}}\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=2(y+\sqrt{y})\cos hx$