Applied Mathematics

Calculus

Medium

applied

If $y = e^{\frac{1}{2}\log(1+ \tan^2 x)}$ , then $\frac{d^2y}{dx^2}$ is equal to:

$\sec x$

$\frac{1}{2}\left(\frac{1}{1+ \tan^2 x}\right)e^{\log(1+ \tan^2 x)}$

$\sec x(\sec^2 x + \tan^2 x)$

$\sec x\tan x$

✅ Correct Option: 3

Given: $y = e^{\frac{1}{2}\log(1+\tan^2 x)}$

Using the power rule of logarithms: $\frac{1}{2}\log(1+\tan^2 x) = \log(1+\tan^2 x)^{1/2}$

Therefore:

$y = e^{\log(1+\tan^2 x)^{1/2}}$

$y = (1+\tan^2 x)^{1/2}$

Using the trigonometric identity $1 + \tan^2 x = \sec^2 x$ :

$y = (\sec^2 x)^{1/2}$

$y = \sec x$

The first derivative:

$\frac{dy}{dx} = \sec x \tan x$

The second derivative using the product rule with $u = \sec x$ and $v = \tan x$ :

$\frac{d^2y}{dx^2} = (\sec x)' \cdot \tan x + \sec x \cdot (\tan x)'$

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

$\frac{d^2y}{dx^2} = \sec x \tan^2 x + \sec^3 x$

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

Therefore, $\frac{d^2y}{dx^2} = \sec x(\sec^2 x + \tan^2 x)$

Solution figure for CUET Mathematics 2025 2 June Shift 1 question 52 (Applied Mathematics)

This is a past-year question by NTA. It has trigonometry, yes. However, it was asked in the Applied Math section.

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(C) Let $C=C(x)$ be the total cost of producing and marketing $x$ units of a product, then marginal cost, $MC=\frac{dC}{dx}$ .

(D) If AC is the average cost, then $AC=\frac{x}{C}$ .

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Choose the correct answer from the options given below:

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The minimum value of $f(x) = 4x^3 - 48x + 105$ in the interval [1,3] is :

If y=e12log⁡(1+tan⁡2x)y = e^{\frac{1}{2}\log(1+ \tan^2 x)}y=e21​log(1+tan2x), then d2ydx2\frac{d^2y}{dx^2}dx2d2y​ is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $y = e^{\frac{1}{2}\log(1+ \tan^2 x)}$ , then $\frac{d^2y}{dx^2}$ is equal to: