1.

Differentiate (cos⁡3x)^3x with respect to x.(a) (cos⁡3x)^x (3 log⁡(cos⁡3x) – 9x tan⁡3x)(b) (cos⁡3x)^3x (3 log⁡(cos⁡3x) + 9x tan⁡3x)(c) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)(d) (cos⁡3x)^3x (log⁡(cos⁡3x) + 9 tan⁡3x)I had been asked this question during an interview.My question comes from Logarithmic Differentiation in division Continuity and Differentiability of Mathematics – Class 12

Answer»

Correct option is (C) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

The best explanation: CONSIDER y=(cos⁡3x)^3x

Applying log on both sides, we get

log⁡y=log⁡(cos⁡3x)^3x

log⁡y=3x log⁡(cos⁡3x)

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{DY}{dx}=\frac{d}{dx} (3x \,log⁡(cos⁡3x))\)

By USING u.v=u’ v+uv’, we get

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{d}{dx} \,(3x) \,log⁡(cos⁡3x)+\frac{d}{dx} \,(log⁡(cos⁡3x)).3x\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \frac{d}{dx} \,(cos⁡3x).3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).\frac{d}{dx}(3x).3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).3.3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) – 9x tan⁡3x)

\(\frac{dy}{dx}\)=(cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)



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