If y=logcosTan-1ex-e-x2 then dydx=

A
$- tanhx$
B
$tanhx$
C
$cothx$
D
$coshx$

Solution:

$\cos \{{Tan}^{- 1} (\frac{e^{x} - e^{- x}}{2})\} = \cos θ, w h e r e θ = {Tan}^{- 1} (\frac{e^{x} - e^{- x}}{2}) = {Tan}^{- 1} (\sin h x) \Rightarrow \tan θ = \sin h x \Rightarrow \cos θ = \frac{1}{\sqrt{1 + \tan^{2} θ}} = \frac{1}{\sqrt{1 + \sin h^{2} x}} = \frac{1}{\sqrt{\cos h^{2} x}} = s e c h x$
$y = \log [\cos \{{Tan}^{- 1} (\frac{e^{x} - e^{- x}}{2})\}] = \log (s e c h x) \Rightarrow \frac{d y}{d x} = \frac{1}{s e c h x} (- s e c h x . \tan h x) = - \tan h x$

If $y = \log [\cos \{{Tan}^{- 1} (\frac{e^{x} - e^{- x}}{2})\}]$ then $\frac{dy}{dx} =$

Solution:

Related content