Question

If $\underset{0}{\overset{y}{\int }}\cos t^{2}dt=\underset{0}{\overset{x^2}{\int }}\frac{\sin t}{t}dt$, then the value of $\frac{dy}{dx}$ is
(where $y=f\left(x\right)$ is a differentiable function)

• A. $\frac{2\sin x}{x\cos y^2}$
• B. $\frac{2\sin x^2}{x\cos y}$
• C. $\frac{2\sin x^2}{x\cos y^2}$
• D. $\frac{2\sin x}{x\cos y}$

Answer 1

The correct option is C $\frac{2\sin x^2}{x\cos y^2}$

Given, $\underset{0}{\overset{y}{\int }}\cos t^{2}dt=\underset{0}{\overset{x^2}{\int }}\frac{\sin t}{t}dt$
Differentiating with respect to $x$ using Leibnitz theorem, we have
$\cos y^2\frac{dy}{dx}=\frac{\sin x^2}{x^2}2x$
$\Rightarrow \frac{dy}{dx}=\frac{2\sin x^2}{x\cos y^2}$