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If $f(x)=\log (\sec x+\tan x)$, then $f^{\prime}\left(\frac{\pi}{4}\right)=$
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Verified Answer
The correct answer is:
$\sqrt{2}$
(B)
$y=\log (\sec x+\tan x)$
Differentially w.r.t. $x$
$\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sec \mathrm{x}+\tan \mathrm{x}} \cdot\left(\sec \mathrm{x} \cdot \tan \mathrm{x}+\sec ^{2} \mathrm{x}\right)=\frac{\sec \mathrm{x}(\tan \mathrm{x}+\sec \mathrm{x})}{(\sec \mathrm{x}+\tan \mathrm{x})} \\
\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{x} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\sec \mathrm{x} \\
\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=\sec \frac{\pi}{4}=\sqrt{2}
\end{array}$
$y=\log (\sec x+\tan x)$
Differentially w.r.t. $x$
$\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sec \mathrm{x}+\tan \mathrm{x}} \cdot\left(\sec \mathrm{x} \cdot \tan \mathrm{x}+\sec ^{2} \mathrm{x}\right)=\frac{\sec \mathrm{x}(\tan \mathrm{x}+\sec \mathrm{x})}{(\sec \mathrm{x}+\tan \mathrm{x})} \\
\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{x} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\sec \mathrm{x} \\
\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=\sec \frac{\pi}{4}=\sqrt{2}
\end{array}$
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