If $f(x)=\log (\sec x+ an x)$, then $f^{\prime}\left(\frac{\pi}{4} ight)=$

Question

If $f(x)=\log (\sec x+	an x)$, then $f^{\prime}\left(\frac{\pi}{4}ight)=$

Aicharya · Accepted Answer

$y=\log (\sec x+	an x)$ Differentially w.r.t. $x$ $\begin{array}{l} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sec \mathrm{x}+	an \mathrm{x}} \cdot\left(\sec \mathrm{x} \cdot 	an \mathrm{x}+\sec ^{2} \mathrm{x}ight)=\frac{\sec \mathrm{x}(	an \mathrm{x}+\sec \mathrm{x})}{(\sec \mathrm{x}+	an \mathrm{x})} \ \frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{x} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\sec \mathrm{x} \ \mathrm{f}^{\prime}\left(\frac{\pi}{4}ight)=\sec \frac{\pi}{4}=\sqrt{2} \end{array}$

If $f(x)=\log (\sec x+\tan x)$, then $f^{\prime}\left(\frac{\pi}{4}\right)=$

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