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Differentiate 2(tan⁡x)^cot⁡x with respect to x.(a) 2 csc^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))(b) csc^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))(c) 2 csc^2⁡x.tan⁡x^cot⁡x (1+log⁡(tan⁡x))(d) 2tan⁡x^cot⁡x (1-log⁡(tan⁡x))The question was asked in a job interview.This intriguing question originated from Logarithmic Differentiation topic in division Continuity and Differentiability of Mathematics – Class 12

Answer» CORRECT choice is (a) 2 CSC^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))

Easy explanation: CONSIDER y=2(tan⁡x)^cot⁡x

Applying log in both sides,

log⁡y=log⁡2(tan⁡x)^cot⁡x

log⁡y=log⁡2+log⁡(tan⁡x)^cot⁡x

log⁡y=log⁡2+cot⁡x log⁡(tan⁡x)

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{DY}{dx}=0+\frac{d}{dx} \,(cot⁡x) \,log⁡(tan⁡x)+cot⁡x \frac{d}{dx} \,(log⁡(tan⁡x))\)

\(\frac{1}{y} \frac{dy}{dx}=-csc^{2⁡}x.log⁡(tan⁡x)+cot⁡x.\frac{1}{tan⁡x}.sec^{2⁡}x\)

\(\frac{dy}{dx} = y\left(-csc^{2⁡x}.log⁡(tan⁡x)+\frac{(1+tan^{2⁡x})}{tan^{2⁡x}}\right)\)

\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x \(\left (-csc^{2⁡x} log⁡(tan⁡x)+cot^{2⁡x}+1 \right )\)

\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x \((-csc^{2⁡x} log⁡(tan⁡x)+csc^{2⁡x})\)

\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x (csc^2⁡x (1-log⁡(tan⁡x))

∴\(\frac{dy}{dx}\)=2 csc^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))


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