Question

Describe the derivatives of each of the piecewise-defined functions in Exercises 45–48.

$f\left(x\right)=\left\{\frac{1}{2}-\frac{1}{4}x,x\le 0;\frac{1}{1+e^x},x>0\right\}$

Short answer

The derivative of function is

$$\begin{gathered} f\text{'}\left(x\right)=-\frac{1}{4},x<0 \\ -\frac{1}{4},x=0 \\ \frac{-e^x}{{\left(1+e^x\right)}^2},x>0 \end{gathered}$$

Step-by-step solution

Step 1. Given Information 

The given function is$f\left(x\right)=\left\{\frac{1}{2}-\frac{1}{4}x,x\le 0;\frac{1}{1+e^x},x>0\right\}$

Step 2. Calculation 

From the differentiation rules,

$$\begin{gathered} \frac{d}{dx}\left(\frac{1}{2}-\frac{1}{4}x\right)=-\frac{1}{4} \\ \frac{d}{dx}\left(\frac{1}{1+e^x}\right)=\frac{-e^x}{{\left(+e^x\right)}^2} \\ \mathrm{let}g\left(x\right)=\frac{1}{2}-\frac{1}{4}x,h\left(x\right)=\frac{1}{1+e^x} \\ \mathrm{And}g\text{'}\left(x\right)=-\frac{1}{4},h\text{'}\left(x\right)=\frac{-e^x}{{\left(+e^x\right)}^2} \end{gathered}$$

Step 3. Calculation

$$\begin{gathered} Checkg\left(0\right)=h\left(0\right) \\ g\left(0\right)=\frac{1}{2} \\ h\left(0\right)=\frac{1}{2} \end{gathered}$$

This is true which follows that the function is continuous at $x=0$

$$\begin{gathered} Checkg\text{'}\left(0\right)=h\text{'}\left(0\right) \\ g\text{'}\left(0\right)=-\frac{1}{4} \\ h\text{'}\left(0\right)=\frac{-e^0}{{\left(1+e^0\right)}^2} \\ =-\frac{1}{4} \end{gathered}$$

So, although function is continuous at zero, it is always differentiable.

Thus, the derivative of the function is

$$\begin{gathered} f\text{'}\left(x\right)=-\frac{1}{4},x<0 \\ -\frac{1}{4},x=0 \\ \frac{-e^x}{{\left(1+e^x\right)}^2},x>0 \end{gathered}$$