Question

Find $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}}^{\mathbf{t}\mathbf{=}\frac{\mathbf{2}}{\mathbf{x}}}\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\mathbf{x}\mathbf{t}}{\mathbf{t}}\mathit{d}\mathit{t}$.

Short answer

The value of$\frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt\text{is}0$ .

Step-by-step solution

Given Information

Given that $\frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt$. ...(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).

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

Solving Further,

$$\begin{gathered} \frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt=\left(\frac{-1}{x}\right)\cosh 2-\left(\frac{-1}{x}\right)\cosh 1+\frac{1}{x}\cosh 2-\frac{1}{x}\cosh 1 \\ \frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt=-\frac{1}{x}\cosh 2+\frac{1}{x}\cosh 1+\frac{1}{x}\cosh 2-\frac{1}{x}\cosh 1 \\ \frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt=0 \end{gathered}$$

Hence the value of $\frac{d}{dx}{\int }_{t=\frac{1}{x}}^{t=\frac{2}{x}}\frac{\cosh xt}{t}dt=0$.