If y=f(cosh x) and f^{prime}(x)=log left(x+sq | vedclass

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(A) Given $y = f(\cosh x)$.Applying the chain rule, we find the first derivative:$\frac{dy}{dx} = f^{\prime}(\cosh x) \cdot \frac{d}{dx}(\cosh x) = f^

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$\sinh x + x \cosh x$

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(A) Given $y = f(\cosh x)$.Applying the chain rule, we find the first derivative:$\frac{dy}{dx} = f^{\prime}(\cosh x) \cdot \frac{d}{dx}(\cosh x) = f^{\prime}(\cosh x) \cdot \sinh x$.Given $f^{\prime}(x) = \log(x + \sqrt{x^2-1})$, we substitute $x$ with $\cosh x$:$f^{\prime}(\cosh x) = \log(\cosh x + \sqrt{\cosh^2 x - 1}) = \log(\cosh x + \sinh x)$.Since $\cosh x + \sinh x = e^x$, we have $f^{\prime}(\cosh x) = \log(e^x) = x$.Thus, $\frac{dy}{dx} = x \sinh x$.Now, differentiate with respect to $x$ using the product rule:$\frac{d^2 y}{dx^2} = \frac{d}{dx}(x \sinh x) = \sinh x + x \cosh x$.

If $y=f(\cosh x)$ and $f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right)$, then $\frac{d^2 y}{d x^2}=$

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