MHT CET · Maths · Differentiation
If \(y=(\sin x)^{\tan x}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is equal to
- A \((\sin x)^{\tan x}\left(1+\sec ^2 x \log (\sin x)\right)\)
- B \(\tan x(\sin x)^{\tan x-1} \cos x\)
- C \((\sin x)^{\tan x} \sec ^2 x \log \sin x\)
- D \(\tan x(\sin x)^{\tan x-1}\)
Answer & Solution
Correct Answer
(A) \((\sin x)^{\tan x}\left(1+\sec ^2 x \log (\sin x)\right)\)
Step-by-step Solution
Detailed explanation
\(y=(\sin x)^{\tan x}\)
Taking logarithm on both sides, we get
\(\log y=\tan x \cdot \log (\sin x)\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{1}{y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan x \cdot \cot x+\log (\sin x) \cdot \sec ^2 x \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=(\sin x)^{\tan x}\left[1+\sec ^2 x \log (\sin x)\right]
\end{aligned}\)
Taking logarithm on both sides, we get
\(\log y=\tan x \cdot \log (\sin x)\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{1}{y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan x \cdot \cot x+\log (\sin x) \cdot \sec ^2 x \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=(\sin x)^{\tan x}\left[1+\sec ^2 x \log (\sin x)\right]
\end{aligned}\)
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