Question

lf $y=\log \cos \left({\tan }^{-1}\right(\frac{e^x-e^{-x}}{2}\left)\right)$ , then $\frac{dy}{dx}=$

• A. $-\tan hx$
• B. $\sin hx$
• C. $\cos hx$
• D. $\cot hx$

Answer 1

The correct option is A $-\tan hx$

$ta{n}^{-1}\left(\frac{e^x-e^{-x}}{2}\right)=co{s}^{-1}\left(\frac{2}{e^x+e^{-x}}\right)$

$y=\log \left(\frac{2}{e^x+e^{-x}}\right)$$=\log 2-\log (e^x+e^{-x})$

$y^{\prime }=-\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)$

$y^{\prime }=-\left(\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}}\right)$

$y^{\prime }=-\frac{sinhx}{coshx}$

$=-tanhx$