Question

Differentiate 7x^(2e^2x) with respect to x.

(a) 14e^2x x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))

(b) 14x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))

(c) 14e^2x x^(2e^2x) (2 log⁡x-\(\frac{1}{x}\))

(d) 14e^2x x^(2e^2x) (log⁡x-\(\frac{1}{x}\))

The question was posed to me at a job interview.

My question comes from Logarithmic Differentiation in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

Correct option is (a) 14e^2x x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))

The explanation is: Consider y=7x^(2e^2x)

log⁡y=log⁡7x^(2e^2x)

log⁡y=log⁡7+log⁡x^(2e^2x)

log⁡y=log⁡7+2e^2x log⁡x

Differentiating with respect to x on both sides, we get

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{d}{dx}\) (log⁡7+2e^2x log⁡x)

\(\frac{1}{y} \frac{dy}{dx}\)=0+\(\frac{d}{dx}\) (2e^2x) log⁡x+\(\frac{d}{dx}\) (log⁡x)2e^2x (using u.v=u’ v+uv’)

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

\(\frac{dy}{dx}\)=y\( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)

\(\frac{dy}{dx}\)=7x^(2e^2x) \( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)

\(\frac{dy}{dx}\)=14e^2x x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))