Question

Differentiate \(e^{4x^5}.2x^{log⁡x^2}\) with respect to x.

(a) \(e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)

(b) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)

(c) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5-log⁡2x^2)\)

(d) \(x^{log⁡x^2 -1} (10x^4+log⁡2x^2)\)

I had been asked this question during an online interview.

The origin of the question is Logarithmic Differentiation topic in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

Right choice is (b) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)

Explanation: Consider y=\(e^{4x^5}+2x^{log⁡x^2}\)

Applying log on both sides, we get

log⁡y=\(log⁡e^{4x^5} \,+ \,log⁡2x^{log⁡x^2}\)

log⁡y=\(4x^5+log⁡x^2 \,. \,log⁡2x\)

log⁡y=\(4x^5+2 \,log⁡x \,log⁡2x\)

Differentiating with respect to x, we get

\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+2(\frac{d}{dx} \,(log⁡x) \,log⁡2x+\frac{d}{dx} \,(log⁡2x) \,log⁡x)\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+2\left (\frac{log⁡2x}{x}+\frac{1}{2x}.2.log⁡x\right )\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x+log⁡x)}{x}\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x^2)}{x}\)

\(\frac{dy}{dx}\)=\(y(20x^4+\frac{2(log⁡2x^2)}{x})\)

\(\frac{dy}{dx}\)=\(e^{4x^5}.2x^{log⁡x^2} (20x^4+\frac{2(log⁡2x^2)}{x})\)

\(\frac{dy}{dx}\)=\(4e^{4x^5}.x^{log⁡x^2 -1} (10x^5+log⁡2x^2)\)