Chapter 2.8, Problem 40E

To determine

To find:The first two derivative of given function and sketch the graph of the function.

Answer to Problem 40E

Thefirst and second derivatives of function $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ are $G^{\prime }\left(x\right)=-e^{-x}\left[\frac{1}{2}\frac{1}{x\sqrt{x}}+\frac{1}{\sqrt{x}}\right]$ , and $G^{″}\left(x\right)=\frac{1}{4}\frac{e^{-x}}{x^{\frac{5}{2}}}\left[3+4x+4x^2\right]$ respectively. The graph of $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ in the interval $0\le x\le 2$ (part AB of the curve) is shown in figure.

Explanation of Solution

Given information:

The given function is $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ , the graph to be shown in the interval $0\le x\le 2$ .

Formulaused:

The derivative of function $y=x^n$ is $\frac{dy}{dx}=n{x}^{n-1}$ and that of $y=e^{ax}$ is $\frac{dy}{dx}=a{e}^{ax}$ .

Calculations:

The given function is $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}=e^{-x}\cdot x^{-\frac{1}{2}}$

The first derivative of the function is

  $G^{\prime }\left(x\right)=e^{-x}\cdot \left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1}+\left(-1\right)e^{-x}\cdot x^{-\frac{1}{2}}=-\frac{1}{2}{e}^{-x}\cdot x^{-\frac{3}{2}}-e^{-x}\cdot x^{-\frac{1}{2}}$   ... (1)

  $\Rightarrow G^{\prime }\left(x\right)=-e^{-x}\left[\frac{1}{2}{x}^{-\frac{3}{2}}+x^{-\frac{1}{2}}\right]=-e^{-x}\left[\frac{1}{2}\frac{1}{x\sqrt{x}}+\frac{1}{\sqrt{x}}\right]$  ... (2)

Now, from result (2), $G^{\prime }\left(x\right)=-\frac{1}{2}{e}^{-x}\left(x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right)$

The second derivative of the function is

  $G^{″}\left(x\right)=-\frac{1}{2}{e}^{-x}\frac{d}{dx}\left[x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right]+\frac{d}{dx}\left(-\frac{1}{2}{e}^{-x}\right)\left(x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right)$

  $\Rightarrow G^{″}\left(x\right)=-\frac{1}{2}{e}^{-x}\left[\left(-\frac{3}{2}\right)x^{-\frac{3}{2}-1}+2\left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1}\right]+\frac{1}{2}{e}^{-x}\left(x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right)$

  $\Rightarrow G^{″}\left(x\right)=\frac{1}{2}{e}^{-x}\left[\frac{3}{2}{x}^{-\frac{5}{2}}+x^{-\frac{3}{2}}\right]+\frac{1}{2}{e}^{-x}\left(x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right)$

  $\Rightarrow G^{″}\left(x\right)=\frac{1}{2}{e}^{-x}\left[\frac{3}{2}{x}^{-\frac{5}{2}}+x^{-\frac{3}{2}}+x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right]=\frac{1}{2}{e}^{-x}\left[\frac{3}{2}{x}^{-\frac{5}{2}}+2x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}\right]$

  $\Rightarrow G^{″}\left(x\right)=\frac{1}{4}{e}^{-x}\left[3x^{-\frac{5}{2}}+4x^{-\frac{3}{2}}+4x^{-\frac{1}{2}}\right]=\frac{1}{4}\frac{e^{-x}}{x^{\frac{5}{2}}}\left[3x^{-\frac{5}{2}+\frac{5}{2}}+4x^{-\frac{3}{2}+\frac{5}{2}}+4x^{-\frac{1}{2}+\frac{5}{2}}\right]$

  $G^{″}\left(x\right)=\frac{1}{4}\frac{e^{-x}}{x^{\frac{5}{2}}}\left[3+4x+4x^2\right]$ .

The graph of function $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ in interval (part AB of the curve) is shown here.

  

Graph and its Interpretation:From result (1) we see that there is no point in the interval $0\le x\le 2$ at which $G^{\prime }\left(x\right)\ne 0$ , hence no horizontal tangent in this interval. Further, in interval $0\le x\le 2$ the second derivative $G^{″}\left(x\right)>0$ , hence the curve must be concave up.

Conclusion:

Thefirst and second derivatives of function $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ are $G^{\prime }\left(x\right)=-e^{-x}\left[\frac{1}{2}\frac{1}{x\sqrt{x}}+\frac{1}{\sqrt{x}}\right]$ , and $G^{″}\left(x\right)=\frac{1}{4}\frac{e^{-x}}{x^{\frac{5}{2}}}\left[3+4x+4x^2\right]$ respectively. The graph of $G\left(x\right)=\frac{e^{-x}}{\sqrt{x}}$ in the interval $0\le x\le 2$ (part AB of the curve) is shown in figure.