## The Definition of a Derivative

Read this section to understand the definition of a derivative. Work through practice problems 1-8.

So far we have emphasized the derivative as the slope of the line tangent to a graph. That interpretation is very visual and useful when examining the graph of a function, and we will continue to use it. Derivatives, however, are used in a wide variety of fields and applications, and some of these fields use other interpretations. The following are a few interpretations of the derivative which are commonly used.

General

Rate of Change $f^{\prime}(x)$ is the rate of change of the function at $x$. If the units for $x$ are years and the units for $f(x)$ are people, then the units for $\frac{\text { df }}{\mathrm{dx}}$ are $\frac{\text { people }}{\text { year }}$, a rate of change in population.

Graphical

Slope $\mathrm{f}^{\prime}(\mathrm{x})$ is the slope of the line tangent to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$

Physical

Velocity If $\mathrm{f}(\mathrm{x})$ is the position of an object at time $\mathrm{x}$, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the velocity of the object at time $\mathrm{x}$. If the units for $x$ are hours and $f(x)$ is distance measured in miles, then the units for $\mathbf{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { miles }}{\text { hour }}$, miles per hour, which is a measure of velocity.

Acceleration If $\mathrm{f}(\mathrm{x})$ is the velocity of an object at time $\mathrm{x}$, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the acceleration of the object at time $\mathrm{x}$. If the units are for $x$ are hours and $f(x)$ has the units $\frac{\text { miles }}{\text { hour }}$, then the units for the acceleration $\mathbf{f}^{\prime}(\mathbf{x})$ $=\frac{\text { df }}{\mathrm{dx}}$ are $\frac{\text { miles/hour }}{\text { hour }}=\frac{\text { miles }}{\text { hour }^{2}}$, miles per hour per hour.

Magnification $f '(x)$ is the magnification factor of the function $f$ for points which are close to $x$If $a$ and $b$ are two points very close to $x$, then the distance between $f(a)$ and $f(b)$ will be close to $f '(x)$ times the original distance between $a$ and $b: f(b) – f(a) ≈ f '(x) ( b – a )$

Business

Marginal Cost If $f(x)$ is the total cost of $x$ objects, then $f '(x)$ is the marginal cost, at a production level of $x$. This marginal cost is approximately the additional cost of making one more object once we have already made $x$ objects. If the units for $x$ are bicycles and the units for $f(x)$ are dollars, then the $\mathbf{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { dollars }}{\text { bicycle }}$, the cost per bicycle.

Marginal Profit If $\mathrm{f}(\mathrm{x})$ is the total profit from producing and selling $\mathrm{x}$ objects, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the marginal profit, the profit to be made from producing and selling one more object.

If the units for $\mathrm{x}$ are bicycles and the units for $\mathrm{f}(\mathrm{x})$ are dollars, then the units for $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { dollars }}{\text { bicycle }}$, dollars per bicycle, which is the profit per bicycle.

In business contexts, the word "marginal" usually means the derivative or rate of change of some quantity.

One of the strengths of calculus is that it provides a unity and economy of ideas among diverse applications. The vocabulary and problems may be different, but the ideas and even the notations of calculus are still useful.

Example 4: A small cork is bobbing up and down, and at time t seconds it is $h(t) = sin(t)$ feet above the mean water level (Fig. 7). Find the height, velocity and acceleration of the cork when $t = 2$ seconds. (Include the proper units for each answer.)

Solution: $h(t) = sin(t)$ represents the height of the cork at any time $t$, so the height of the cork when $\mathrm{t}=2$ is $\mathrm{h}(2)=\sin (2) \approx 0.91$ feet.

The velocity is the derivative of the position, so $v(t)=\frac{d h(t)}{d t}=\frac{d \sin (t)}{d t}=\cos (t)$. The derivative of position is the limit of $(\Delta \mathrm{h}) /(\Delta \mathrm{t})$, so the units are (feet)/(seconds). After 2 seconds the velocity is $\mathrm{v}(2)=\cos (2) \approx-0.42$ feet per second $=-0.42 \mathrm{ft} / \mathrm{s}$.

The acceleration is the derivative of the velocity, so $\mathrm{a}(\mathrm{t})=\frac{\mathrm{d} \mathrm{v}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{d} \cos (\mathrm{t})}{\mathrm{dt}}=-\sin (\mathrm{t})$. The derivative of velocity is the limit of $(\Delta \mathrm{v}) /(\Delta \mathrm{t})$, so the units are (feet/second) / (seconds) or feet/second $^{2}$. After 2 seconds the acceleration is $\mathrm{a}(2)=-\sin (2) \approx-0.91 \mathrm{ft} / \mathrm{s}^{2}$.

Practice 6: Find the height, velocity and acceleration of the cork in the previous example after 1 second?