## How do you calculate the derivative of a function from the definition?

The definition of the derivative of a function *f* (*x*) at a point *x* = *a* was given in Section 11.3,

$$f’\left( a \right) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(a + h) – f(a)}}{h}$$

provided this limit exists. We can adapt this definition to find the derivative of a function by changing the constant *a* to a variable *x*.

*f*(

*x*) is defined as $$f’\left( x \right) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(x + h) – f(x)}}{h}$$

provided the limit exists. The symbols

*f*′(

*x*), are read “

*f*prime of

*x*”.

We can apply this definition in a manner similar to how we applied the definition of derivative of a function at a point.

To find the derivative of *f* (*x*) or *f* ′(*x*),

- Evaluate the function
*f*(*x*) at*x*+*h*to give*f*(*x*+*h*). Simplify this expression as much as possible. - Form and simplify the difference quotient $\frac{{f(x + h) – f(x)}}{h}$.
- Take the limit as
*h*approaches 0 of the simplified difference quotient.

Let’s use this strategy to find the derivatives of several functions.

#### Example 2 Find the Derivative

Use the definition of the derivative to find the derivative of the function $$f(x) = 2x – 7$$

**Solution** To apply the definition of the derivative to this function, we must evaluate *f* (*x* + *h*). Replace *x* in the function with *x* + *h* to yield $$\displaylines{

f(x + h) = 2\left( {x + h} \right) – 7 \cr

= 2x + 2h – 7 \cr} $$

It is important to note that we are replacing *x* with the group *x* + *h* in parentheses. Now form the difference quotient,$$\frac{{f(x + h) – f(x)}}{h} = \frac{{\left( {2x + 2h – 7} \right) – \left( {2x – 7} \right)}}{h}$$

To make the limit easy to evaluate, let’s simplify the difference quotient as much as possible.

$$

\begin{align*}

\frac{{f(x + h) – f(x)}}{h} &= \frac{{\left( {2x + 2h – 7} \right) – \left( {2x – 7} \right)}}{h} && {\color{red}{\small \text{Remove the parentheses and subtract each term}}} \cr

&= \frac{{2x + 2h – 7 \color{red}{- 2x + 7}}}{h} && \cr

&= \frac{{2 \color{red}{h}}}{\color{red}{h}} && {\color{red}{\small \text{Combine like terms}}} \cr

&= 2 && {\color{red}{\small \text{Reduce the quotient}}} \cr

\end{align*} $$

Substitute this expression into the definition of the derivative:

$$\begin{align*}

f'(x) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(x + h) – f(x)}}{h} \cr

&= \mathop {\lim }\limits_{h\,\, \to 0} 2 \cr

&= 2 \cr \end{align*}$$

The derivative of the linear function $f(x) = 2x – 7$ is $f'(x) = 2$.

#### Example 3 Find the Derivative

Use the definition of the derivative to find the derivative of the function$$g(t) = {t^2} + 3t + 7$$

Solution In this example, we’ll follow the same strategy for finding the derivative. However, since the name of the function is *g*(*t*), we need to adjust the definition of the derivative to read$$g'(t) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{g(t + h) – g(t)}}{h}$$

Substitute *t* + *h* for *t* in *g*(*t*) to give

$$

\begin{align*}

g(t + h) &= {\left( {t + h} \right)^2} + 3\left( {t + h} \right) + 7 && \cr

&= {t^2} + 2ht + {h^{^2}} + 3\left( {t + h} \right) + 7 && \color{red}{\small \text{Multiply }} \color{red}{\left( {t+h} \right)\left( {t+h} \right)} \cr

&= {t^2} + 2ht + {h^2} + 3t + 3h + 7 && \color{red}{\small \text{Multiply 3 times }} \color{red}{t+h} \cr

\end{align*}

$$

If we place this expression and the expression for into the numerator of the difference quotient, we get

$$\begin{align*}

\frac{{g(t + h) – g(t)}}{h} &= \frac{{\left( {{t^2} + 2ht + {h^2} + 3t + 3h + 7} \right) – \left( {{t^2} + 3t + 7} \right)}}{h} && \color{red}{\small \text{Subtract and simplify}} \cr

&= \frac{{{t^2} + 2ht + {h^2} + 3t + 3h + 7 \color{red}{- {t^2} – 3t – 7}}}{h} && \color{red}{\small \text{Combine like terms}} \cr

&= \frac{{2ht + {h^2} + 3h}}{h} && \cr

&= \frac{{\color{red}{h}\left( {2t + h + 3} \right)}}{\color{red}{h}} && \color{red}{{\small \text{Factor }}h{\text{ and reduce}}} \cr

&= 2t + h + 3 && \cr

\end{align*}

$$

With this expression in the definition of the derivative, we write

$$

\begin{align*}

g'(t) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{g(t + h) – g(t)}}{h} \cr

&= \mathop {\lim }\limits_{h\,\, \to 0} \left( {2t + h + 3} \right) \cr

&= 2t + 3 \cr

\end{align*}

$$

The derivative of the quadratic function $g(t) = {t^2} + 3t + 7$ is the linear function $g'(t) = 2t + 3$.

#### Example 4 Find the Derivative

Use the definition of the derivative to find the derivative of the function$$j(t) = {e^t}$$

**Solution** The definition of the derivative for this function is$$j'(t) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{j(t + h) – j(t)}}{h}$$

Using the exponential function, the difference quotient is$$\frac{{j(t + h) – j(t)}}{h} = \frac{{{e^{t + h}} – {e^t}}}{h}$$

This expression can be simplified, but not as easily as Example 2 or Example 3.

In this case we rewrite ${e^{t + h}}$ as ${e^t}\;{e^h}$. By doing this, we can factor the numerator:

$$

\begin{align*}

\frac{{j(t + h) – j(t)}}{h} &= \frac{{{e^{t + h}} – {e^t}}}{h} \cr

&= \frac{{{e^t}\;{e^h} – {e^t}}}{h} \cr

&= \frac{{{e^t}\left( {{e^h} – 1} \right)}}{h} \cr

\end{align*}

$$

From this simplified difference quotient, we can write out the definition of the derivative,

$$

\begin{align*}

j'(t) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{j(t + h) – j(t)}}{h} \cr

&= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^t}\left( {{e^h} – 1} \right)}}{h} \cr

&= {e^t} \cdot \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} \cr

\end{align*}

$$

The factor ${e^t}$ is a constant with respect to *h*, so it can be moved outside the limit. To complete the derivative, we must evaluate the limit.

For this limit we’ll find the limit by creating a table of values for for smaller and smaller *h* values.

h |
$\frac{{{e^h} – 1}}{h}$ |

0.1 | 1.051709181 |

0.01 | 1.005016708 |

0.001 | 1.000500167 |

0.0001 | 1.000050002 |

0.00001 | 1.000005000 |

The table suggests that as *x* gets smaller and smaller, the value of the quotient approaches 1 or $$\mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} = 1$$

Using $\mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} = 1$ in the expression for the definition of the derivative yields

$$

\begin{align*}

j'(t) &= {e^t} \cdot \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} \cr

&= {e^t} \cdot 1 \cr

&= {e^t} \cr

\end{align*}

$$

The derivative of $j(t) = {e^t}$ is $j'(t) = {e^t}$.