The derivative as a function
Drawing f′
The derivative f'(a) is the slope of the tangent line at one point. Compute that slope at every point and you get a new function, f'. What does its graph look like, and how could you sketch it without any formula?
Drag the ochre point along the top graph. Each slope you visit is recorded as a point on the bottom graph. Before you sweep across, predict where the bottom graph will cross the axis.
y = f(x), with the tangent line at the ochre point
y = f′(x), traced from the slopes you have visited
Drag the ochre point, or use Sweep. Points on the bottom graph appear only where you have been.
Reading f′ from the shape of f
The derivative function is defined by the same limit as before, now with x in place of a fixed point:
Other common notations are \frac{dy}{dx}, \frac{d}{dx}f(x) and y'. You can sketch f' from the graph of f using four facts.
- f rising
f'>0, so the graph of f' is above the axis. Steeper rising means larger values.
- f falling
f'<0, so the graph of f' is below the axis.
- Horizontal tangent
f'=0. Peaks and valleys of f become zeros of f'.
- Corner or vertical tangent
f' does not exist there. At a corner it jumps. At a vertical tangent it grows without bound. Try |x| and \sqrt[3]{x^2}.
A function that is differentiable at a point is also continuous there. The reverse is false: |x| is continuous at 0 but has no derivative there.
Solve it step by step
Read the graph of f and work out what you can about f′, then pick its graph. The score counts steps you answer correctly on the first try.
Common mistakes
- Copying the height of f
f′ records slope, not height. Where f is high but flat, f′ is 0.
- Putting f′'s zeros at f's zeros
The zeros of f′ sit under the peaks and valleys of f, not where f crosses the axis.
- Smoothing over a corner
At a corner of f, the graph of f′ has a gap. Do not join the two sides.