If y = log(cosh x),then frac{d^2 y}{d x^2} = | vedclass

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(A) Given $y = \log(\cosh x)$.First,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \frac{d}{dx}(\cosh x)$Since $\frac{d}{d

Vedclass · Accepted Answer

$\operatorname{sech}^2 x$

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(A) Given $y = \log(\cosh x)$.First,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \frac{d}{dx}(\cosh x)$Since $\frac{d}{dx}(\cosh x) = \sinh x$,we get:$\frac{dy}{dx} = \frac{\sinh x}{\cosh x} = 	anh x$Now,differentiate again with respect to $x$:$\frac{d^2 y}{dx^2} = \frac{d}{dx}(	anh x)$Since $\frac{d}{dx}(	anh x) = \operatorname{sech}^2 x$,we have:$\frac{d^2 y}{dx^2} = \operatorname{sech}^2 x$Thus,the correct option is $A$.

If $y = \log(\cosh x)$,then $\frac{d^2 y}{d x^2} = $

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