Question

What is the value of \(\frac{d}{dx}\)(e^x sinx + e^x cos ⁡x)?

(a) 0

(b) 2 cos⁡x

(c) 2e^x.sin ⁡x

(d) 2e^x.cos⁡ x

The question was posed to me during an interview.

This intriguing question originated from Derivatives topic in section Limits and Derivatives of Mathematics – Class 11

Answer 1

Right choice is (d) 2e^x.cos⁡ x

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)

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

\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) =(e^x.cos⁡ x + sin ⁡x . e^x) + (e^x.(-sin ⁡x) + cos ⁡x.e^x)

\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = e^x.cos⁡ x + sin⁡ x . e^x – e^x.sin⁡ x + cos ⁡x.e^x

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