MHT CET · Maths · Differentiation
The derivative of \(f(\tan x)\) w.r.t. \(g(\sec x)\) at \(x=\frac{\pi}{4}\), where \(f^{\prime}(1)=2\) and \(g^{\prime}(\sqrt{2})=4\) is
- A \(\frac{1}{\sqrt{2}}\)
- B 2
- C \(\sqrt{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\frac{\frac{d}{d x} f(\tan x)}{\frac{d}{d x} g(\sec x)}=\frac{f^{\prime}(\tan x) \cdot \sec ^{2} x}{g^{\prime}(\sec x) \cdot \sec x \tan x} \\
\text { At } x=\frac{\pi}{4} \text { we get } \\
=\frac{f^{\prime}\left(\tan \frac{\pi}{4}\right)\left(\sec ^{2} \frac{\pi}{4}\right)}{g^{\prime}\left(\sec \frac{\pi}{4}\right) \sec \frac{\pi}{4} \tan \frac{\pi}{4}}=\frac{f^{\prime}(1)(2)}{g^{\prime}(\sqrt{2})(\sqrt{2})(1)}=\frac{2 \times 2}{4 \sqrt{2}}=\frac{1}{\sqrt{2}}
\end{array}
\)
\begin{array}{l}
\frac{\frac{d}{d x} f(\tan x)}{\frac{d}{d x} g(\sec x)}=\frac{f^{\prime}(\tan x) \cdot \sec ^{2} x}{g^{\prime}(\sec x) \cdot \sec x \tan x} \\
\text { At } x=\frac{\pi}{4} \text { we get } \\
=\frac{f^{\prime}\left(\tan \frac{\pi}{4}\right)\left(\sec ^{2} \frac{\pi}{4}\right)}{g^{\prime}\left(\sec \frac{\pi}{4}\right) \sec \frac{\pi}{4} \tan \frac{\pi}{4}}=\frac{f^{\prime}(1)(2)}{g^{\prime}(\sqrt{2})(\sqrt{2})(1)}=\frac{2 \times 2}{4 \sqrt{2}}=\frac{1}{\sqrt{2}}
\end{array}
\)
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