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Differentiate (log⁡2x)^sin⁡3x with respect to x.(a) (3 cos⁡3x log⁡(log⁡2x)+\(\frac{sin⁡3x}{x log⁡2x}\))(b) \(log⁡2x^{sin⁡3x} \,(3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x})\)(c) –\((3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x log⁡2x})\)(d) \(\frac{3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x log⁡2x}}{log⁡2x^{sin⁡3x}}\)The question was posed to me during an interview.The origin of the question is Logarithmic Differentiation topic in portion Continuity and Differentiability of Mathematics – Class 12

Answer»

Correct choice is (B) \(LOG⁡2x^{sin⁡3x} \,(3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{X \,log⁡2x})\)

The explanation is: Consider y=\((log⁡2x)^{sin⁡3x}\)

Applying log on both sides, we get

log⁡y=\(log⁡(log⁡2x)^{sin⁡3x}\)

log⁡y=sin⁡3x log⁡(log⁡2x)

Differentiating with RESPECT to x, we get

\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x)\frac{d}{dx} (sin⁡3x)+sin⁡3x \frac{d}{dx} \,(log⁡(log⁡2x))\)

By using chain rule, we get

\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x).3 \,cos⁡3x+sin⁡3x.\frac{1}{log⁡2x}.\frac{1}{2x}.2 \,(∵u.v=u’ \,v+uv’)\)

\(\frac{dy}{dx}\)=y(3 cos⁡3x log⁡(log⁡2x)+\(\frac{sin⁡3x}{x \,log⁡2x}\))

∴\(\frac{dy}{dx}\)=log⁡2x^sin⁡3x \(\left (3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x} \right )\)


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