Question

If y = x + tan x, then value of \(\frac{d^2y}{dx^2}\) at \(x= \frac{\pi}{4}\) is:

• A. 1
• B. 2
• C. 4
• D. \(2 \sqrt{2}\)

Answer 1

The Correct Option is C

To find the second derivative of \(y = x + \tan x\) and evaluate it at \(x = \frac{\pi}{4}\), follow these steps:

First, calculate the first derivative \( \frac{dy}{dx} \):

\(\frac{dy}{dx} = \frac{d}{dx}(x + \tan x) = 1 + \sec^2 x\)

Next, find the second derivative \( \frac{d^2y}{dx^2} \):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(1 + \sec^2 x) = 0 + 2\sec^2 x \tan x\)

Now, substitute \(x = \frac{\pi}{4}\) into the second derivative:

\(\sec\left(\frac{\pi}{4}\right) = \sqrt{2}\) and \(\tan\left(\frac{\pi}{4}\right) = 1\)

Thus, \(\frac{d^2y}{dx^2} \Big|_{x=\frac{\pi}{4}} = 2(\sqrt{2})^2 \cdot 1 = 2 \times 2 \times 1 = 4\)

Therefore, the value of \(\frac{d^2y}{dx^2}\) at \(x = \frac{\pi}{4}\) is 4.