Question

If $\mathit{y}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}{\mathit{t}}^{\mathbf{2}}\mathbf{d}\mathit{t}$, find $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$.

Short answer

The value of$\frac{dy}{dx}$ is $\frac{\sin x}{2\sqrt{x}}$.

Step-by-step solution

Given Information

Given that $y={\int }_0^{\sqrt{x}}\sin t^{2}dt$. ...(1)

Formula Used

We know that$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathbf{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathbf{d}\mathit{t}$ ...(2)

Find dydx

Compare equation (1) and equation (2).

$f=\sin \left(t^2\right),u=0\mathrm{and}v=\sqrt{x}$

Substitute values of f, u and v in the equation (2).

$$\begin{gathered} \frac{dy}{dx}=f\left(\sqrt{x}\right)\frac{d\sqrt{x}}{dt}-f\left(0\right)\frac{d\left(0\right)}{dx} \\ =\sin x\times \frac{1}{2}{x}^{-\frac{1}{2}} \\ =\frac{\sin x}{2\sqrt{x}} \end{gathered}$$

Hence the value of$\frac{dy}{dx}$ is $\frac{\sin x}{2\sqrt{x}}$.