Using the Definition

Let f(x) = x² + 1 and find f'(x).

$$\begin{eqnarray*}f'(x) &=& \displaystyle\lim_{\bigtriangleup x \to 0} \displayst... ...yle\lim_{\bigtriangleup x \to 0} 2x + \bigtriangleup x\\ &=& 2x \end{eqnarray*}$$

Since f'(x) = 2x we can find the slope of the function for any x (or the rate of change of the function). So when x=3 the derivative is equal to

f'(3) = 2(3) = 6

A Non-Differentiable Function at x=0

Let $f(x) = \mid x \mid$. Then

$$f(x) = \left\{ \begin{array}{r@{\quad:\quad}l} x & x\ge 0\\ -x & x < 0\\ \end{array} \right.$$

Find f'(0) using the definition:

$$\begin{eqnarray*}f'(0)^{+} &=& \displaystyle\lim_{\bigtriangleup x \to 0^{+}} \d... ...splaystyle\frac{\bigtriangleup x }{\bigtriangleup x}\\ &=& 1\\ \end{eqnarray*}$$

$$\begin{eqnarray*}f'(0)^{-} &=& \displaystyle\lim_{\bigtriangleup x \to 0^{-}} \d... ...aystyle\frac{- \bigtriangleup x }{\bigtriangleup x}\\ &=& -1\\ \end{eqnarray*}$$

Notice that the limit from the left and the limit from the right are not equal. Differentiation is dependent on this limit existing. Because the left handed and right handed limits are not equal the limit does not exist, and thus the function is not differentiable at x=0.


Other functions which are not differentiable at x=0 are f(x) = x^2/3 or f(x) = x^1/3. The first is similar to the absolute value function in that it takes a sharp turn at x=0. The second has an unbounded derivative at x=0. Check for yourself. Where do you think the function $f(x) = \mid x-4 \mid$ is not differentiable? Think of where it takes a sharp turn.