Chapter 2.8, Problem 17E

To determine

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

Answer to Problem 17E

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

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

Explanation of Solution

Given:

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

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 $F\left(z\right)=\frac{1+e^{-z}}{1+e^z}$, use the quotient rule and chain rule of derivatives and simplify further as shown below:

  $F\text{'}\left(z\right)=\frac{d}{dz}\left(\frac{1+e^{-z}}{1+e^z}\right)$

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

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

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

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

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

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

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

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

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

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

  $F\text{'}\text{'}\left(z\right)=\frac{d}{dz}\left(\frac{-e^{-z}-e^z-2}{{\left(1+e^z\right)}^2}\right)$

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

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

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

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

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

  $=\frac{\left[\left(e^{-z}-e^z+1-e^{2z}\right)+\left(2+2e^{2z}+4e^z\right)\right]}{{\left(1+e^z\right)}^3}$

  $=\frac{\left[e^{-z}-e^z+1-e^{2z}+2+2e^{2z}+4e^z\right]}{{\left(1+e^z\right)}^3}$

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

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