Chapter 2.8, Problem 18E

To determine

The firstand second derivativesof the function, $p\left(x\right)=\frac{1-e^x}{1+e^x}$ .

Answer to Problem 18E

  $p\text{'}\left(x\right)=\frac{-2e^x}{{\left(1+e^x\right)}^2}$ ,

  $p\text{'}\text{'}\left(x\right)=\frac{2e^x\left(e^x-1\right)}{{\left(1+e^x\right)}^3}$

Explanation of Solution

Given:

The function, $p\left(x\right)=\frac{1-e^x}{1+e^x}$ .

Concept Used:

• The quotient rule of derivatives:

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\cdot \frac{d}{dx}f\left(x\right)-f\left(x\right)\cdot \frac{d}{dx}g\left(x\right)}{{\left(g\left(x\right)\right)}^2}$

• The product rule of derivatives:

$\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]=f\left(x\right)\cdot \frac{d}{dx}g\left(x\right)+g\left(x\right)\cdot \frac{d}{dx}f\left(x\right)$

• Chain rule, $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)\cdot g\text{'}\left(x\right)$
• Power rule of derivatives, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$
• $\frac{d}{dx}\left(e^x\right)=e^x$
• Derivative of any constant C is zero, that is, $\frac{d}{dx}\left(C\right)=0$

Calculation:

In order to find the derivatives of the given function $p\left(x\right)=\frac{1-e^x}{1+e^x}$ , use the quotient rule and chain rule of derivatives and simplify further as shown below:

  $\begin{array}{c}p\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{1-e^x}{1+e^x}\right)\\ =\frac{\left(1+e^x\right)\cdot \frac{d}{dx}\left(1-e^x\right)-\left(1-e^x\right)\cdot \frac{d}{dx}\left(1+e^x\right)}{{\left(1+e^x\right)}^2}\\ =\frac{\left(1+e^x\right)\cdot \left[\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(e^x\right)\right]-\left(1-e^x\right)\cdot \left[\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(e^x\right)\right]}{{\left(1+e^x\right)}^2}\\ =\frac{\left(1+e^x\right)\cdot \left[\left(0\right)-\left(e^x\right)\right]-\left(1-e^x\right)\cdot \left[\left(0\right)+\left(e^x\right)\right]}{{\left(1+e^x\right)}^2}\\ =\frac{\left(1+e^x\right)\cdot \left[-e^x\right]-\left(1-e^x\right)e^x}{{\left(1+e^x\right)}^2}\\ =\frac{-e^x-e^{2x}-e^x+e^{2x}}{{\left(1+e^x\right)}^2}\\ =\frac{-2e^x}{{\left(1+e^x\right)}^2}\end{array}$

So, the first derivative is $p\text{'}\left(x\right)=\frac{-2e^x}{{\left(1+e^x\right)}^2}$ .

Now, to find the second derivative of the function, use the quotient ruleof derivatives and simplify further using the chain rule as shown below

  $p\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{-2e^x}{{\left(1+e^x\right)}^2}\right)$

  $=\frac{{\left(1+e^x\right)}^2\cdot \frac{d}{dx}\left(-2e^x\right)-\left(-2e^x\right)\cdot \frac{d}{dx}{\left(1+e^x\right)}^2}{{\left({\left(1+e^x\right)}^2\right)}^2}$

  $=\frac{{\left(1+e^x\right)}^2\cdot \left[-2\frac{d}{dx}\left(e^x\right)\right]+2e^x\cdot \left[2{\left(1+e^x\right)}^{2-1}\cdot \frac{d}{dx}\left(1+e^x\right)\right]}{{\left(1+e^x\right)}^4}$

  $=\frac{{\left(1+e^x\right)}^2\cdot \left[-2\left(e^x\right)\right]+2e^x\cdot \left[2\left(1+e^x\right)\cdot \left(0+e^x\right)\right]}{{\left(1+e^x\right)}^4}$

  $=\frac{-2e^x{\left(1+e^x\right)}^2+2e^x\cdot \left[2e^x\left(1+e^x\right)\right]}{{\left(1+e^x\right)}^4}$

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

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

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

Thus, the second derivative of the given function is $p\text{'}\text{'}\left(x\right)=\frac{2e^x\left(e^x-1\right)}{{\left(1+e^x\right)}^3}$ .