Question

Differentiate 3 sin^-1⁡(e^2x) w.r.t x.

(a) \(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)

(b) \(\frac{2e^2x}{\sqrt{1-e^{4x}}}\)

(c) –\(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)

(d) \(\frac{6e^{-2x}}{\sqrt{1-e^{4x}}}\)

This question was addressed to me by my school principal while I was bunking the class.

This interesting question is from Exponential and Logarithmic Functions in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct choice is (a) \(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)

Explanation: Consider y=3 sin^-1⁡(e^2x)

\(\frac{dy}{dx}=\frac{d}{dx}\)(3 sin^-1⁡(e^2x))

\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\frac{d}{dx}\)(e^2x)

\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\)2e^2x

∴\(\frac{dy}{dx}\)=\(\frac{6e^{2x}}{\sqrt{1-e^{4x}}}\)