MHT CET · Maths · Differentiation
If \(f(x)=\log (\sec x+\tan x)\), then \(f^{\prime}\left(\frac{\pi}{4}\right)=\)
- A 1
- B \(\frac{2}{\sqrt{3}}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
(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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