Question

Find$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$ explicitly if $\mathit{y}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\frac{{\mathbf{e}}^{\mathbf{x}\mathbf{u}}\mathbf{-}\mathbf{1}}{\mathbf{u}}\mathit{d}\mathit{u}$.

Short answer

The value of $\frac{dy}{dx}$is $\frac{e^x-1}{x}$.

Step-by-step solution

Given Information

Given that $y={\int }_0^1\frac{{\mathrm{e}}^{\mathrm{xu}}-1}{\mathrm{u}}du$. ...(1)

Formula Used

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

Finding the value of dydx

Compare equation (1) and equation (2).

$$\begin{gathered} \frac{d}{dx}\left(y\right)=\frac{d}{dx}{\int }_0^1\frac{e^{xu}-1}{u}du \\ \frac{dy}{dx}=\frac{e^{x\left(1\right)}-1}{1}·\frac{d}{dx}\left(1\right)-\frac{e^{x\left(0\right)}-1}{1}·\frac{d}{dx}\left(0\right)+{\int }_0^{1}u\frac{e^{xu}}{u}du \\ \frac{dy}{dx}=\frac{e^{x\left(1\right)}-1}{1}\left(0\right)-\frac{e^{x\left(0\right)}-1}{1}\left(0\right)+{\int }_0^1{e}^{xu}du \\ \frac{dy}{dx}={\int }_0^1{e}^{xu}du \end{gathered}$$

Solving further,

$$\begin{gathered} \frac{dy}{dx}={\left[\frac{e^{xu}}{x}\right]}_0^1 \\ =\frac{1}{x}\left(e^{x\left(1\right)}-e^{x\left(0\right)}\right) \\ =\frac{e^x-1}{x} \end{gathered}$$

Therefore, $\frac{dy}{dx}=\frac{e^x-1}{x}$.