Question

Evaluate: $\frac{d}{dx}\left(\frac{e^{\sin \sqrt{x}}+e^{-\sin \sqrt{x}}}{e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}}\right)$

• A. $\frac{\frac{-2\cos \sqrt{x}}{\sqrt{x}}}{(e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}{)}^2}$
• B. $\frac{\frac{-2\cos \sqrt{x}}{\sqrt{x}}}{e^{\sin \sqrt{x}}}$
• C. $\frac{-2\cos \sqrt{x}}{\sin \sqrt{x}}$
• D. $\frac{2\cos \sqrt{x}}{e^{\sin \sqrt{x}}}$

Answer 1

The correct option is A $\frac{\frac{-2\cos \sqrt{x}}{\sqrt{x}}}{(e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}{)}^2}$

$\frac{e^{\sin \sqrt{x}}+e^{-\sin \sqrt{x}}}{e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}}$

$=\frac{\frac{e^{\sin \sqrt{x}}+e^{-\sin \sqrt{x}}}{2}}{\frac{e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}}{2}}$

$=\frac{\cosh \left(\sin \sqrt{x}\right)}{\sinh \left(\sin \sqrt{x}\right)}=\coth \left(\sin \sqrt{x}\right)$.

Now,

$\frac{d}{dx}\left[\frac{e^{\sin \sqrt{x}}+e^{-\sin \sqrt{x}}}{e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}}}\right]=\frac{d}{dx}\left(\text{coth}\sin \sqrt{x}\right)=-{\text{cosech}}^2\left(\sin \sqrt{x}\right).\left(\cos \sqrt{x}\right).\frac{1}{2\sqrt{x}}=\frac{-\frac{4\cos \sqrt{x}}{2\sqrt{x}}}{{(e^{sin\sqrt{x}}-e^{-\sin \sqrt{x}})}^2}$

$=\frac{-\frac{2\cos \sqrt{x}}{\sqrt{x}}}{{(e^{\sin \sqrt{x}}-e^{-\sin \sqrt{x}})}^2}$