Question

Differentiate 8e^cos2x w.r.t x.

(a) 16 sin⁡2x e^cos2x

(b) -16 sin⁡2x e^cos2x

(c) -16 sin⁡2x e^-cos⁡2x

(d) 16 sin⁡2x e^-cos⁡2x

I had been asked this question in an interview.

The doubt is from Exponential and Logarithmic Functions in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct option is (b) -16 sin⁡2x e^cos2x

Best explanation: Consider y=8e^cos2x

Differentiating w.r.t x by using chain rule, we get

\(\frac{dy}{dx}=\frac{d}{dx}\)(8e^cos2x)

=8e^cos2x\(\frac{d}{dx}\) (cos⁡2x)

=8e^cos2x(-sin⁡2x)\(\frac{d}{dx}\) (2x)

=8e^cos2x(-sin⁡2x)(2)

∴\(\frac{dy}{dx}\)=-16 sin⁡2x e^cos2x