Tangent Lines and Derivatives: Videos & Practice Problems

Understanding the difference between secant and tangent lines is crucial in calculus. A secant line intersects a curve at two points, while a tangent line touches the curve at just one point. To find the slope of a tangent line, we use the limit as x approaches a specific value, expressed as $lim\left(f\right(x)-f(c\left)\right)/(x-c))$. This process leads to the derivative, representing the instantaneous rate of change at a point.

Slopes of Tangent Lines

So at this point of the course, we should be familiar with something known as the secant line, and recall that the secant line is just a line that intersects the curve of our function at two points. Let's say we have this curve down here of the function $f(x)=x^2$, we can draw a secant line, which intersects this curve at two specific points. Now in this video, we're going to be learning all about this thing known as the tangent line. While there are some similarities between the secant line and the tangent line, there are also some very key differences, and understanding these differences is going to be very important in this course. So without further ado, let's just jump right into an example to see what these types of problems would look like.

In this example here, we're told given the function $f(x)=x^2$, find the slope of the tangent line at $x=1$. Right off the bat, there's already a key difference I noticed with this type of problem. Notice how we are only given one value of interest, one point, $x=1$. For secant lines, we had two values that we were given because there were two points on our graph, but here, we're only given one. And that actually is the key difference between a secant line and a tangent line.

Where a secant line intersects the curve at two points, a tangent line touches it will barely touch the curve at one point. To understand this, let's go ahead and take a look at these differences down here where we can see that we have this secant line. Imagine for the secant line, we take one of these points here, and we just bring it very close to the other point, so it approaches that value, and in essence, we would only be looking at one point on the graph. We can visualize this by taking this secant line and having this point approach the value of $x=1$, having the two points converge very close together.

This is really the main idea of a tangent line. What we could do is only write one point right here at $x=1$, and we could go ahead and draw a line that is tangent to that point. This right here would be the tangent line, and finding the slope of this line would allow us to solve this problem. Now the question, of course, becomes how do we calculate this slope? Well, it turns out it's a very similar process to how we calculated a secant line, except since we only have one point that is approaching the other point, recall this language of approaching we've discussed before when dealing with limits.

So rather than having two points, we're only going to be looking at one point of interest, except we're going to have our entire function approach this limit, where we have the limit as $x$ approaches this point of interest. So this right here is the equation for finding the slope of the tangent line. Notice how we keep our function and an $x$ value in there because we're approaching getting very close to that value that we're looking at. Let's go ahead and see if we can calculate the slope of this tangent line. $c$ is the value of interest, which I can see up here is 1.

What this means is that our $c$ value is 1. I also see that our function is $x^2$. So if I want to find the slope of this tangent line, it is going to be the limit as $x$ approaches 1 of our function $f(x)-f(1)$ all divided by $x-1$, and this again is the $c$ value. Now to do this, I'll go ahead and take our function and plug it in for $f(x)$. So what that means is we're going to have the function $x^2-f(1)$, and $f(1)$ is just going to be the function evaluated at 1.

We're going to have $f(1)$, which is going to be $1^2$, and that's simply equal to 1. Now all of this is going to be divided by $x-1$. At this point, what I can do is try applying this limit. But notice if I take this $x$ approaching 1 here and I plug it in for $x$, we're going to get $1^2-1$ divided by $1-1$, which is all just going to become 0 over 0. This is something that we're not allowed to have in mathematics.

So what we need to do is find a way to simplify this and somehow get it so we're not dividing by 0. What I can do is try factoring the top of this fraction, because $x^2-1$ is the same thing as $(x-1)(x+1)$. This is the difference of squares, and all of this is going to be divided by $x-1$. Now from here, I can cancel the $x-1$, so all I'm going to have is the limit as $x$ approaches 1 for $x+1$. Now at this point, I can take this value of 1 and plug it in for $x$. So we're going to have $1+1$, which is equal to 2. So, 2 is the solution to this problem, and that is the slope of this tangent line.

Notice that we got a slope of 2 when we had the tangent line, but we got a slope of 4 when we had the secant line. That's because we have a steeper slope here since we're looking at two points on the graph, rather than just one point. Now there is one more thing I want to mention, another difference between the secant and tangent line. When calculating the slope of a secant line, we call this the average rate of change. And I think that makes sense since we're, in essence, averaging these two points on the graph.

But if we're looking at the slope of a tangent line, we actually call this the instantaneous rate of change since we're looking at the slope, or the rate of change at just that one value, that one point, so it's instantaneous. Now we also give this a third name, which is called the derivative. So maybe you've heard of derivatives before, but basically, a derivative is just the slope of a tangent line at one point. So this is how you can solve these types of problems, where you need to calculate the derivative, AKA the instantaneous rate of change, AKA the slope of a tangent line. So I hope you found this video helpful, and let's try getting some more practice.

Given the function $f(x)=4x^2-1$, calculate the slope of the tangent line at $x=-3$.

• A. 36
• B. -36
• C. -24
• D. 24

Answer: (C)

Given the function $f(x)=x^2-10x+2$, calculate the slope of the tangent line at $x=2$.

• A. -6
• B. 6
• C. 10
• D. -10

Answer: (A)

Given the function $f(x)=x^2+100$, calculate the slope of the tangent line at $x=0$.

• A. 0
• B. 100
• C. -100
• D. Infinite (vertical line)

Answer: (A)

Equations of Tangent Lines

So, in a recent video, we learned all about tangent lines and how to calculate the slope of a tangent line using this equation. Now recall that this is just a line that touches the curve of your function at exactly one point; that's what a tangent line is. In this video, we're going to now learn how to solve some other problems associated with tangent lines, specifically finding their equations. If this sounds scary, don't sweat it, because it turns out the process for doing this is actually related to concepts that we've already learned up to this point. So, without further ado, let's just jump right into an example to see what these kinds of problems look like.

So here, we're asked to find the equation of the line tangent to this function, $f(x)=3x^2-4$ at $x=-2$. So how can we go about solving this? Well, what I'm going to do is do this by the steps, and our first step is going to be to plug our $x$ value of interest, which we call $c$, we discussed that in the previous video, into our function $f(x)$ to get our $y$ value $f(c)$. Now, I can see here that our $c$ value is going to be -2 since that's the x-value of interest that we're looking at. So since our $x$ value is -2, what we need to do is figure out what $f(c)$ is, which is going to be $f(-2)$.

And $f(-2)$, while plugging it into this function, will have $3\times (-2{)}^2-4$. Now $3\times (-2{)}^2$ turns out to be 12, and 12 minus 4 is 8. So what we end up with is 8 for our $f(c)$ value. So now that we've found $c$ and $f(c)$, what we can do is move on to step 2, which is plugging things into this equation for the slope of the tangent line. We already figured out that the $c$ here is -2, so this is going to stay consistent for this limit.

But as for the rest of the equation, well, we need to plug our function in for $f(x)$. $f(x)$, we can see, is $3x^2-4$. Now this whole thing is going to be minus $f(c)$, which we just calculated to be 8. And all of this will be divided by $x-c$, which is the same thing as $x-(-2)$ or $x+2$. So this is what we get for our equation.

Now from here, what I'm going to do is simplify what we have on top since I can see I can take this -4 and subtract off 8. That's going to give me -12, so we're going to have $3x^2-12$ all over $x+2$. Now at this point, what I'm going to do is try applying my limit by pulling -2 in for $x$. But I noticed that if I do that, I'm going to end up with -2 + 2 on the bottom, which is 0. You cannot divide by 0.

You cannot have 0 in the denominator of a fraction. So we're going to need to find some sort of other strategy here to evaluate this limit. Now what I'm going to do is take this 3 and factor it out of both these values, because I can see that 3 would go on the outside, leaving us with just $x^2$ there, and 12 divided by 3 is 4. So this whole thing will become $x^2-4$, and then all of that will be divided by $x+2$. Now I still have this $x+2$ in the bottom, so I can't plug my -2 in yet.

But what I notice is that we have $x^2-4$ on top. $x^2-4$ is a difference of squares, because we could also write this as $x^2-2^2$, because 2 squared is 4, all divided by $x+2$. Since we have this difference in squares, this whole thing could be expanded out to be $(x+2)(x-2)$. And, again, this is all divided by $x+2$. Now from here, the $x+2$s are going to cancel, meaning all that I'm going to have is $3(x-2)$.

Now notice this $x+2$ in the denominator went away, so I can now plug my value of -2 in. So what we're going to have is $3\times (-2-2)$. Now -2 - 2 is -4, and 3 times -4 is -12. So that there is the slope of the tangent line. So notice that we've gone ahead and evaluated this limit, and we did have to factor and cancel.

But that's very common when dealing with these types of limits. You're going to have to do that step. So now that we found what our slope of the tangent line is, we now need to figure out what the equation of our tangent line is. But we can actually do this because notice that we have an $x$ and $y$ value up here, and we have a slope. When we have these values, we can just use point-slope form.

So that's the strategy for solving these types of problems. You first want to calculate the slope, and then you want to plug everything into point-slope form to get your equation. So let's just go ahead and do this. So we have $y-$ our $y$ value, which is going to be $y-$ this value of 8. And then it's going to be equal to our slope, which we just calculated to be -12.

This was our slope. And then it's going to be times $x$ minus our $x$ value, which is -2, or we could just write this as $x+2$, because minus a negative would be plus. So now that we have everything plugged into point-slope form, our last step is going to be to solve this equation for $y$. So let's go ahead and do that. Now the first thing I'm going to do is distribute this -12 into the parenthesis.

So what we're going to end up with is -12 times $x$, which is -12$x$, and then -12 times 2, which is -24. Now from here, I'm going to add 8 on both sides of this equation, which will get the -8 and the positive 8 to cancel on the left. So that will give me $y$ by itself, which is what I'm looking for, and then we'll have -12$x$, and then -24 + 8 is -16. So $y=-12x-16$ is the solution to this problem, and that's how you can find the equation of the tangent line. So this is how you can solve these types of problems.

Hope you found this video helpful, and let's try getting some more practice with this.

Given the function $f(x)=3(x^2-1)$, find the equation of the tangent line at $x=1$.

• A. $y=-6x+6$
• B. $y=-6x-6$
• C. $y=6x+6$
• D. $y=6x-6$

Answer: (D)

Determine Continuity Algebraically Example 1

Welcome back, everyone. So let's go ahead and try solving this example. In this problem, we're told to find the equation of the line tangent to the function f(x)=-4x2 at x=-2. To start with this problem, we need the equation for the slope of the tangent line, which is the limit as x approaches c of our function f(x)-f(c) all divided by x-c. This is the general equation for the slope of a tangent line.

I can apply the function and the numbers that were given in this equation. c is going to be our given x value, which is -2, so we'll have the limit as x approaches -2. Our function f(x) is -4x2 as given.

We have -4x2 minus f(c). Now f(-2) is going to be, using this equation, -4×42 which is -16. So that means we're going to have -4x2 minus -16 all divided by x--2. We can simplify to -4x2+16 divided by x+2.

Now, when attempting to apply the limit by plugging -2 for x, we notice that the denominator becomes zero, a situation we cannot have in mathematics, hence requiring a different strategy. Thus, we factor the numerator, extracting -4, resulting in x2-4 all divided by x+2. Recognizing the numerator as a difference of squares, we rewrite it, leading to cancellation in the denominator and simplify to -4×(x-2).

Applying the limit, the slope of our tangent line at x=-2 becomes 16. With this slope, to find the equation of the tangent line, use point-slope form: y-y00=m×(x-x00), using the point (-2,-16). Simplifying, we find the tangent line’s equation: y=16x+16.

Hope you found this video helpful.

Here’s what students ask on this topic:

What is the difference between a secant line and a tangent line in calculus?

A secant line intersects a curve at two points, while a tangent line touches the curve at just one point. The secant line represents the average rate of change between two points on the curve, whereas the tangent line represents the instantaneous rate of change at a single point. The slope of the tangent line is found using the limit as x approaches a specific value, expressed as $\frac{lim\left(f\right(x)-f(c\left)\right)}{(x-c)}$. This process leads to the derivative, representing the instantaneous rate of change at a point.

How do you find the slope of a tangent line at a given point?

To find the slope of a tangent line at a given point, you use the limit definition of the derivative. For a function $f\left(x\right)$, the slope of the tangent line at $x=c$ is given by $\frac{lim\left(f\right(x)-f(c\left)\right)}{(x-c)}$ as $x$ approaches $c$. This limit, if it exists, is the derivative of the function at that point, representing the instantaneous rate of change.

What is the equation of a tangent line to a function at a given point?

To find the equation of a tangent line to a function $f\left(x\right)$ at a given point $(c,f(c\left)\right)$, follow these steps: 1) Find the slope of the tangent line using the derivative $f\text{'}\left(c\right)$. 2) Use the point-slope form of a line: $y-f\left(c\right)=f\text{'}\left(c\right)(x-c)$. This equation represents the tangent line at the point $(c,f(c\left)\right)$.

What is the instantaneous rate of change and how is it related to the derivative?

The instantaneous rate of change of a function at a specific point is the rate at which the function's value is changing at that exact point. It is represented by the slope of the tangent line to the function at that point. This concept is mathematically expressed as the derivative of the function. The derivative, denoted as $f\text{'}\left(x\right)$, is found using the limit definition: $\frac{lim\left(f\right(x)-f(c\left)\right)}{(x-c)}$ as $x$ approaches $c$. This derivative gives the instantaneous rate of change at the point $(c,f(c\left)\right)$.

How do you find the derivative of a function using the limit definition?

To find the derivative of a function $f\left(x\right)$ using the limit definition, follow these steps: 1) Identify the function $f\left(x\right)$ and the point of interest $c$. 2) Set up the limit expression: $\frac{lim\left(f\right(x)-f(c\left)\right)}{(x-c)}$ as $x$ approaches $c$. 3) Simplify the expression to eliminate any indeterminate forms (e.g., 0/0). 4) Evaluate the limit to find the derivative $f\text{'}\left(c\right)$. This process gives the instantaneous rate of change of the function at the point $(c,f(c\left)\right)$.