The derivative at a point

Example 1: Let $f(x)=x^2-8x+9$. Find $f'(3)$, and the equation of the tangent line at $x=3$.

Solution:

Step 1: $f(3)=3^2-8(3)+9=-6$.

Step 2: Expand $f(3+h)$:

$\begin{align*}&f(3+h)\\&=(3+h)^2-8(3+h)+9\\&=9+6h+h^2-24-8h+9\\&=-6-2h+h^2\end{align*}$

Step 3: First find the numerator of the difference quotient:

$\begin{align*}&f(3+h)-f(3)\\&=(-6-2h+h^2)-(-6)\\&=-2h+h^2\end{align*}$

Step 4: Divide by $h$:

$\begin{align*}&f'(3)\\&=\lim_{h\to 0}\dfrac{-2h+h^2}{h}\end{align*}$

Cancel the common factor $h$:

$\begin{align*}&=\lim_{h\to 0}(-2+h)\end{align*}$

Let $h$ approach $0$. Put $0$ in for $h$:

$\begin{align*}&f'(3)=-2+0=-2\end{align*}$

The tangent line goes through $(3,-6)$ with slope $-2$:

$\begin{align*}y-(-6)&=-2(x-3)\\y+6&=-2x+6\\y&=-2x\end{align*}$

Example 2: Let $f(x)=x^3$. Find $f'(2)$.

Solution:

Here $f(2)=8$. Use the definition of the derivative:

$\begin{align*}&f'(2)\\&=\lim_{h\to 0}\dfrac{(2+h)^3-8}{h}\end{align*}$

Expand

$(2+h)^3=8+12h+6h^2+h^3$.

The $8$s cancel:

$\begin{align*}&f'(2)\\&=\lim_{h\to 0}\dfrac{12h+6h^2+h^3}{h}\end{align*}$

Cancel the common factor $h$:

$\begin{align*}&=\lim_{h\to 0}(12+6h+h^2)\end{align*}$

Let $h$ approach $0$. Put $0$ in for $h$:

$\begin{align*}&f'(2)=12+6(0)+0^2=12\end{align*}$

Example 3: Let $f(x)=\dfrac{1}{x+1}$. Find $f'(1)$.

Solution:

Here $f(1)=\tfrac{1}{2}$ and $f(1+h)=\tfrac{1}{2+h}$. Use the definition of the derivative:

$\begin{align*}&f'(1)\\&=\lim_{h\to 0}\dfrac{\dfrac{1}{2+h}-\dfrac{1}{2}}{h}\end{align*}$

Combine the two fractions in the numerator, using the common denominator $2(2+h)$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{\dfrac{2-(2+h)}{2(2+h)}}{h}\end{align*}$

Simplify the numerator. Dividing by $h$ puts $h$ in the denominator:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-h}{2h(2+h)}\end{align*}$

Cancel the common factor $h$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{-1}{2(2+h)}\end{align*}$

Let $h$ approach $0$. Put $0$ in for $h$:

$\begin{align*}&f'(1)=\dfrac{-1}{2(2+0)}=-\dfrac{1}{4}\end{align*}$

Example 4: Let $f(x)=\sqrt{x}$. Use the ${x\to a}$ form to find $f'(9)$.

Solution:

Here $f(9)=3$. Use the $x\to a$ form of the derivative:

$\begin{align*}&f'(9)\\&=\lim_{x\to 9}\dfrac{\sqrt{x}-3}{x-9}\end{align*}$

The denominator is a difference of squares:

$x-9=(\sqrt{x}-3)(\sqrt{x}+3)$.

So

$\begin{align*}&f'(9)\\&=\lim_{x\to 9}\dfrac{\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\end{align*}$

Cancel the common factor $\sqrt{x}-3$:

$\begin{align*}&=\lim_{x\to 9}\dfrac{1}{\sqrt{x}+3}\end{align*}$

Let $x$ approach $9$. Put $9$ in for $x$:

$\begin{align*}&f'(9)=\dfrac{1}{\sqrt{9}+3}=\dfrac{1}{6}\end{align*}$

Example 5: The graph shows a function $f$ and its tangent line at $x=2$. Find $f'(2)$.

Solution:

The tangent line goes through $(2,1)$ and $(4,3)$. From $(2,1)$ to $(4,3)$, the run is $2$ and the rise is $2$. So,

$f'(2)=\dfrac{\text{rise}}{\text{run}}=\dfrac{2}{2}=1$.

Example 6: Water is draining from a tank. $V(t)$ is the volume of water in the tank, in liters, $t$ minutes after the draining starts. What does $V'(10)=-4$ mean?

Solution:

$V'(10)$ is the rate of change of the volume at $t=10$ minutes. Its units are liters per minute. The negative sign means the volume is decreasing.

So, $10$ minutes after the draining starts, the water is leaving the tank at a rate of $4$ liters per minute.

Example 7: Let $f(x)=|x|$. Show that $f'(0)$ does not exist.

Solution:

Here $f(0)=0$, so the difference quotient is $\dfrac{|h|}{h}$. In one-sided limits (Example 4), we found:

$\displaystyle\lim_{h\to 0^-}\dfrac{|h|}{h}=-1$ and $\displaystyle\lim_{h\to 0^+}\dfrac{|h|}{h}=1$.

The one-sided limits are different, so the limit does not exist. So $f'(0)$ does not exist, and $f(x)=|x|$ is not differentiable at $0$.

On the graph, $y=|x|$ has a sharp corner at $(0,0)$. The slope is $-1$ on the left and $1$ on the right, so there is no single tangent line there. We look at this more closely in the next section.