Question

Find \(\frac{d^2y}{dx^2}\), if y=tan^2⁡x+3 tan⁡x.

(a) sec^2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)

(b) 2 sec^2⁡⁡x tan⁡x (2 tan⁡x-sec⁡x+3)

(c) 2 sec^2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)

(d) 2 sec^2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x-3)

This question was posed to me in an interview.

The above asked question is from Second Order Derivatives topic in section Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct answer is (c) 2 sec^2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)

Explanation: Given that, y=tan^2⁡⁡x+3 tan⁡x

\(\frac{dy}{dx}\)=2 tan⁡x sec^2⁡⁡x+3 sec^2⁡x=sec^2⁡⁡x (2 tan⁡x+3)

By using the u.v rule, we get

\(\frac{d^2 y}{dx^2}=\frac{d}{dx}\) (sec^2⁡⁡x).(2 tan⁡x+3)+\(\frac{d}{dx}\) (2 tan⁡x+3).sec^2⁡⁡x

\(\frac{d^2 y}{dx^2}\)=2 sec^2⁡⁡x tan⁡x (2 tan⁡x+3)+sec^2⁡⁡x (2 sec⁡x tanx)

=2 sec^2⁡x tan⁡x (2 tan⁡x+sec⁡x+3).