Question

Differentiate 4x^e^x with respect to x.

(a) x^e^x e^-x (x log⁡x+1)

(b) -4x^e^x-1 e^x (x log⁡x+1)

(c) 4x^e^x e^x (x log⁡x+1)

(d) 4x^e^x-1 e^x (x log⁡x+1)

The question was posed to me during a job interview.

Asked question is from Logarithmic Differentiation in portion Continuity and Differentiability of Mathematics – Class 12

Answer 1

Right option is (d) 4x^e^x-1 e^x (x log⁡x+1)

Explanation: Consider y=4x^e^x

Applying log on both sides, we get

log⁡y=log⁡4x^e^x

log⁡y=log⁡4+log⁡x^e^x (∵log⁡ab=log⁡a+log⁡b)

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{dy}{dx}=0+\frac{d}{dx}(e^x \,log⁡x)(∵log⁡a^b=b \,log⁡a)\)

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

\(\frac{dy}{dx}=y(e^x log⁡x+\frac{e^x}{x})\)

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