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### Course: Class 11 Physics (India)>Unit 4

Lesson 6: Using the formal definition of derivative

# Tangent lines and rates of change

How tangent lines are a limit of secant lines, and where the derivative and rate of change fit into all this.

## Introduction

The position of a car driving down the street, the value of currency adjusted for inflation, the number of bacteria in a culture, and the AC voltage of an electric signal are all examples of quantities that change with time. In this section, we will study the rate of change of a quantity and how is it related geometrically to secant and tangent lines.

## Secant and tangent lines

If two distinct points $P\left({x}_{0},{y}_{0}\right)$ and $Q\left({x}_{1},{y}_{1}\right)$ lie on the curve $y=f\left(x\right)$, the slope of the secant line connecting the two points is
${m}_{\mathrm{sec}}=\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}=\frac{f\left({x}_{1}\right)-f\left({x}_{0}\right)}{{x}_{1}-{x}_{0}}\phantom{\rule{0.167em}{0ex}}$.
If we let the point ${x}_{1}$ approach ${x}_{0}$, then $Q$ will approach $P$ along the graph $f$. The slope of the secant line through points $P$ and $Q$ will gradually approach the slope of the tangent line through $P$ as ${x}_{1}$ approaches ${x}_{0}$. In the limit, the previous equation becomes
${m}_{\mathrm{tan}}=\underset{{x}_{1}\to {x}_{0}}{lim}\frac{f\left({x}_{1}\right)-f\left({x}_{0}\right)}{{x}_{1}-{x}_{0}}\phantom{\rule{0.167em}{0ex}}$.
If we let $h={x}_{1}-{x}_{0}$, then ${x}_{1}={x}_{0}+h$ and $h\to 0$ as ${x}_{1}\to {x}_{0}$. We can rewrite the limit as
${m}_{\mathrm{tan}}=\underset{h\to 0}{lim}\frac{f\left({x}_{0}+h\right)-f\left({x}_{0}\right)}{h}$.
When the limit exists, its value ${m}_{\mathrm{tan}}$ is the slope of the tangent line to the graph of $f$ at the point $P\left({x}_{0},{y}_{0}\right)$.

## Example 1

Find the slope of the tangent line to the graph of the function $f\left(x\right)={x}^{3}$ at the point $\left(2,8\right)$.

## Solution

Since $\left({x}_{0},{y}_{0}\right)=\left(2,8\right)$, using the slope of the tangent line formula
${m}_{\mathrm{tan}}=\underset{h\to 0}{lim}\frac{f\left({x}_{0}+h\right)-f\left({x}_{0}\right)}{h}$
we get
$\begin{array}{rl}{m}_{\mathrm{tan}}& =\underset{h\to 0}{lim}\frac{f\left(2+h\right)-f\left(2\right)}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{\left({h}^{3}+6{h}^{2}+12h+8\right)-8}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{{h}^{3}+6{h}^{2}+12h}{h}\\ \\ & =\underset{h\to 0}{lim}\left({h}^{2}+6h+12\right)\\ \\ & =12.\end{array}$
Thus, the slope of the tangent line is $12$. Recall from algebra that the point-slope form of the equation of the tangent line is
$y-{y}_{0}={m}_{\mathrm{tan}}\cdot \left(x-{x}_{0}\right)$.
The point-slope formula gives us the equation
$y-8=12\cdot \left(x-2\right)$
which we can rewrite as
$y=12x-16$.

## Finding the slope at any point

Next we are interested in finding a formula for the slope of the tangent line at any point on the graph of $f$. Such a formula would be the same formula that we are using except we replace the constant ${x}_{0}$ by the variable $x$. This yields
${m}_{\mathrm{tan}}=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}\phantom{\rule{0.167em}{0ex}}$.
We denote this formula by
$f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}\phantom{\rule{0.167em}{0ex}}$,
where $f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)$ is read "$f$ prime of $x$." The next example illustrates its usefulness.

## Example 2

If $f\left(x\right)={x}^{2}-3$, find $f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)$ and use the result to find the slopes of the tangent lines at $x=2$ and $x=-1$.

## Solution

Since
$f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}$,
then
$\begin{array}{rl}f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)& =\underset{h\to 0}{lim}\frac{\left[\left(x+h{\right)}^{2}-3\right]-\left[{x}^{2}-3\right]}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{{x}^{2}+2xh+{h}^{2}-3-{x}^{2}+3}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{2xh+{h}^{2}}{h}\\ \\ & =\underset{h\to 0}{lim}\left(2x+h\right)\\ \\ & =2x.\end{array}$
To find the slope, we substitute $x=2$ and $x=-1$ into the result $f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)$. We get
$f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(2\right)=2\left(2\right)=4$
and
$f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(-1\right)=2\left(-1\right)=-2$.
Thus, slopes of the tangent lines at $x=2$ and $x=-1$ are $4$ and $-2$, respectively.

## Example 3

Find the slope of the tangent line to the graph of $f\left(x\right)=1/x$ at the point $\left(1,1\right)$.

## Solution

Using the slope of the tangent line formula
$f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
and substituting $f\left(x\right)=1/x$ gives us
$\begin{array}{rl}{f}^{\prime }\left(x\right)& =\underset{h\to 0}{lim}\frac{\left(\frac{1}{x+h}\right)-\frac{1}{x}}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{\frac{x-\left(x+h\right)}{x\left(x+h\right)}}{h}\\ \\ & =\underset{h\to 0}{lim}\frac{x-x-h}{hx\left(x+h\right)}\\ \\ & =\underset{h\to 0}{lim}\frac{-h}{hx\left(x+h\right)}\\ \\ & =\underset{h\to 0}{lim}\frac{-1}{x\left(x+h\right)}\\ \\ & =\frac{-1}{{x}^{2}}.\end{array}$
Substituting $x=1$ yields
${f}^{\prime }\left(1\right)=\frac{-1}{\left(1{\right)}^{2}}=-1$.
Thus, the slope of the tangent line at $x=1$ for the graph of $f\left(x\right)=1/x$ is $m=-1$. To find the equation of the tangent line, we use the point-slope formula,
$y-{y}_{0}=m\cdot \left(x-{x}_{0}\right)$,
where $\left({x}_{0},{y}_{0}\right)=\left(1,1\right)$. The equation of the tangent line is
$\begin{array}{rl}y-1& =-1\cdot \left(x-1\right)\\ y& =-x+1+1\\ y& =-x+2.\end{array}$

## Average speed

The primary concept of differential is calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of change of the distance traveled with respect to time. If a car travels $120$ miles in $4$ hours, his speed is
.
This speed is called the average speed or the average rate of change of distance with respect to time. Of course, a car that travels $120$ miles at an average rate of $30$ miles per hour for $4$ hours does not necessarily do so at constant speed. It may have slowed down or sped up during the $4$ hour period.
However, if the car hits a tree, it would not be its average speed that determines the resulting damage but its speed at the instant of the collision. So here we have two distinct kinds of speeds, average speed and instantaneous speed.
The average speed of an object is defined as the object’s displacement $\mathrm{△}x$ divided by the time interval $\mathrm{△}t$ during which the displacement occurs:
$v=\frac{\mathrm{△}x}{\mathrm{△}t}=\frac{{x}_{1}-{x}_{0}}{{t}_{1}-{t}_{0}}$
The average speed is also the expression for the slope of a secant line connecting the two points. Figure 1 shows the $\text{secant line}$ through the points $\left({t}_{0},{x}_{0}\right)$ and $\left({t}_{1},{x}_{1}\right)$ on the $\text{position-versus-time curve}$.
Thus we conclude that the average speed of an object between time ${t}_{0}$ and ${t}_{1}$ is represented geometrically by the slope of the secant line connecting the two points $\left({t}_{0},{x}_{0}\right)$ and $\left({t}_{1},{x}_{1}\right)$. If we choose ${t}_{1}$ close to ${t}_{0}$, then the average speed will closely approximate the instantaneous speed at time ${t}_{0}$.

## Rates of change

The average rate of change of an arbitrary function $f$ on an interval is represented geometrically by the slope of the secant line to the graph of $f$. The instantaneous rate of change of $f$ at a particular point is represented by the slope of the tangent line to the graph of $f$ at that point. Let's consider each case in more detail.

### Average rate of change

The average rate of change of the function $f$ over the interval $\left[{x}_{0},{x}_{1}\right]$ is
${m}_{\mathrm{sec}}=\frac{f\left({x}_{1}\right)-f\left({x}_{0}\right)}{{x}_{1}-{x}_{0}}$
Figure 2 shows the $\text{secant line}$ through the points $\left({x}_{0},f\left({x}_{0}\right)\right)$ and $\left({x}_{1},f\left({x}_{1}\right)\right)$ on the $\text{graph of}\phantom{\rule{0.167em}{0ex}}f$. The slope of the secant line is the average rate of change ${m}_{\mathrm{sec}}$.

### Instantaneous rate of change

The instantaneous rate of change of the function $f$ at the point ${x}_{0}$ is
${m}_{\mathrm{tan}}=f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left({x}_{0}\right)=\underset{{x}_{1}\to {x}_{0}}{lim}\frac{f\left({x}_{1}\right)-f\left({x}_{0}\right)}{{x}_{1}-{x}_{0}}$
Figure 3 shows $\text{tangent line}$ through the point $\left({x}_{0},f\left({x}_{0}\right)\right)$ on the $\text{graph of}\phantom{\rule{0.167em}{0ex}}f$. The slope of the tangent line is the instantaneous rate of change ${m}_{\mathrm{tan}}$.

## Example 4

Suppose that $y={x}^{2}-3$.
(a) Find the average rate of change of $y$ with respect to $x$ over the interval $\left[0,2\right]$.
(b) Find the instantaneous rate of change of $y$ with respect to $x$ at the point $x=-1$.

## Solution

(a) Applying the formula for average rate of change with $f\left(x\right)={x}^{2}-3$ and ${x}_{0}=0$ and ${x}_{1}=2$ yields
$\begin{array}{rl}{m}_{\mathrm{sec}}& =\frac{f\left({x}_{1}\right)-f\left({x}_{0}\right)}{{x}_{1}-{x}_{0}}\\ \\ & =\frac{f\left(2\right)-f\left(0\right)}{2-0}\\ \\ & =\frac{1-\left(-3\right)}{2}\\ \\ & =2\end{array}$
This means the average rate of change over the interval $\left[0,2\right]$ is 2 units of increase in $y$ for each unit of increase in $x$.
(b) From Example 2 above, we found that $f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(x\right)=2x$, so
$\begin{array}{rl}{m}_{\mathrm{tan}}& =f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left({x}_{0}\right)\\ & =f{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(-1\right)\\ & =2\left(-1\right)\\ & =-2.\end{array}$
This means that the instantaneous rate of change is negative. That is, $y$ is decreasing at $x=-1$. It is decreasing at a rate of $2$ units in $y$ for each unit of increase in $x$.

## Want to join the conversation?

• What is the purpose of the function that appears at the very end of example 2? f(x)= 7 | x=1, f(x-1)+5 | x>1
What is this related to?
• Do I understand correctly that the growth "h" is at the end of reduction of limit omited because its value is so small it has practically no effect on value of slope?
• Yes, because technically speaking we write "as the limit of h approaches zero" which means h keeps getting smaller and smaller until it is negligible.
• can someone explain to the math for substituting y=1/x in Example 3?
for some reason i cant see what happens in step one and two
• How did we find the function at the end of example 2? f(x)=7 and f(x)=f(x-1)+5?
• It seems to be an orphan that was not removed before the lesson was posted. It doesn't relate to the previous example and is not related to the next example.
• In the example 1, I have not idea how do they get h^3+6h^2+12h+8 from f(2+h)−f(2), it is really confuse me. Can anyone explain?
• They're simply evaluating f(x) for x = 2 + h. (2 + h)^3 is equal to h^3 + 6*h^2 + 12*h + 8.
• In figure 1, it doesn't make a lot of sense using a position vs time graph if you're talking about speed and that specific graph wasn't really a good choice since its slope can be negative and speed cannot be negative. You should use a displacement vs time graph instead or it's a bit confusing.
• I agree with Juan. I think that rather than using y to represent displacement (as it does in figure 1), a monotonically increasing function should be used to represent distance, since after all, we are looking at the rate of change in distance with respect to time.
(1 vote)
• On example 3, when taking the derivative of 1/x,
at the third step while using the formula - "h" is somehow transferred from the lower denominator to the denominator above it:
x-(x+h)/x(x+h)/h --> x-(x+h)/xh(x+h)
Which algebric rule relates to that? I couldn't figure this out.
• This is a complex fraction simplified. ((x - (x-h)) / x(x+h)) / h is the same thing as (x - (x-h)) / x(x+h)) * 1/h, in other words a/b/c = a/b * 1/c = a/bc