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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⁡ xThe 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»

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


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