If we have the graph of $$y=f(x)$$, we can draw a sketch of the graph of $$y=f'(x)$$ by estimating the slope of the tangent to the graph of $$f$$ at each $$x$$ value. We then plot the points $$(x,f'(x))$$ in the $$xy$$-plane and connect them by a smooth curve whenever possible. This curve represents the graph of $$y=f'(x)$$.

When you seek to graph the derivative, it is often easiest to first identify the places where the slope of the tangent line is 0. Those are the places where you can hope the sign on the derivative may change.

Example 1
The graph of$$y=f(x)$$ is shown below. Give a rough sketch of the graph of $$y=f'(x).$$

Solution

The tangent to the curve is horizontal at $$x=0$$ and $$x=2.$$
$\begin{cases} x<0 & \text{tangent line makes an acute angle with positive } x \text{ axis}\Rightarrow f'(x)>0\\ 0<x<2 & \text{tangent line makes an obtuse angle with positive }x\text{ axis}\Rightarrow f'(x)<0\\ 2<x & \text{tangent line makes an acute angle with positive }x \text{ axis}\Rightarrow f'(x)>0 \end{cases}$
Specifically, the slope of the tangent line at $$x=-1$$ is
$m_{\text{tan}}=\frac{3}{1}\Rightarrow f'(-1)=3.$
The slope of the tangent line at $$x=1$$ is
$m_{\text{tan}}=\frac{-1}{1}\Rightarrow f'(1)=-1$
At $$x=3$$
$m_{\text{tan}}=\frac{-3}{-1}\Rightarrow f'(3)=3.$
If we connects these points:
$(-1,3),(0,0),(1,-1),(2,0),(3,3)$ we can figure out how $$f’$$ looks like (Figure 1).