Question

Find $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{1}}^{\frac{\mathbf{1}}{\mathbf{x}}}\frac{{\mathbf{e}}^{\mathbf{x}\mathbf{t}}}{\mathbf{t}}\mathit{d}\mathit{t}$.

Short answer

The value of$\frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt\text{is}-\frac{e^x}{x}$ .

Step-by-step solution

Given Information

Given that . $\frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt$ ...(1)

Formula Used

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

Solving the given equation

Compare equation (1) and equation (2).

$$\begin{gathered} \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=\frac{e^{x\left(\frac{1}{x}\right)}}{\frac{1}{x}}\times \frac{d}{dx}\left(\frac{1}{x}\right)-\frac{e^{x\left(1\right)}}{1}\times \frac{d}{dx}\left(1\right)+{\int }_1^{\frac{1}{x}}\frac{t{e}^{xt}}{t}dt \\ \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=x{e}^1\left(\frac{-1}{x^2}\right)-\frac{e^x}{1}\left(0\right)+{\int }_1^{\frac{1}{x}}{e}^{xt}dt \\ \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=-\frac{1}{x}{e}^1+{\left[\frac{e^{xt}}{x}\right]}_1^{\frac{1}{x}} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=-\frac{1}{x}{e}^1+\left[\frac{e^{x\left(\frac{1}{x}\right)}}{x}-\frac{e^{x\left(1\right)}}{x}\right] \\ \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=-\frac{1}{x}{e}^1+e^1\frac{1}{x}-\frac{e^x}{x} \\ \frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=-\frac{e^x}{x} \end{gathered}$$

Hence the value of$\frac{d}{dx}{\int }_1^{\frac{1}{x}}\frac{e^{xt}}{t}dt=-\frac{e^x}{x}$ .