Question

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

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

Answer 1

Answer: (A)

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