Question

What is the value of \(\frac{d}{dx}\) (e^x tan x) at x = 0?

(a) 0

(b) 1

(c) -1

(d) 2

The question was posed to me in class test.

The doubt is from Derivatives in division Limits and Derivatives of Mathematics – Class 11

Answer 1

Correct answer is (b) 1

Easy explanation: We need to use product rule in both the terms to get the answer.

\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)

Here f = e^x and g = tan ⁡x

\(\frac{d}{dx}\) (e^x tan x) = tan⁡ x.\(\frac{d}{dx}\) (e^x) + e^x.\(\frac{d}{dx}\) (tan ⁡x)

\(\frac{d}{dx}\) (e^x tan x) = tan⁡ x.e^x + e^x . sec^2⁡x

At x = 0 we get,

= tan ⁡0.e^0 + e^0.sec^2⁡0

= 0.(1) + 1.(1)

= 1