Tangent Lines & Derivatives - Online Tutor, Practice Problems & Exam Prep

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, representing the instantaneous rate of change, or derivative. The derivative can be calculated using the limit as \( h \) approaches 0: $lim\left(f\right(x+h)-f(x\left)\right)/h$. This allows for finding the slope of the tangent line at any point on a function.

Slopes of Tangent Lines

So at this point in the course, we should be familiar with something known as the secant line, and recall that the secant line is just a line which intersects the curve of our function at two points. So let's say we have this curve down here of the function x2, we can draw a secant line, which intersects this curve at 2 specific points. Now in this video, we're going to be learning all about this thing known as the tangent line. And 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. So in this example here, we're told given the function fx=x2, find the slope of the tangent line at x=1. Now 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 2 values that we were given because there were two points on our graph, but here, we're only given 1. And that actually is the key difference between a secant line and a tangent line. Where a secant line intersects the curve at 2 points, a tangent line touches. It only will barely touch the curve at one point.

So to understand this, well, 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 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. So we can visualize this by taking this secant line and having this point approach this value of x=1, having the two points converge very close together. So this is really the main idea of a tangent line. So what we could do is we can 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 exactly do we calculate this slope? Well, it turns out it's actually 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 2 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. And 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. So let's go ahead and see if we can calculate the slope of this tangent line. Now c is the value of interest, which I can see up here is 1. So what this means is that our c value is 1. I also see that our function is x2. 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 fx minus f1 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 fx. So what that means is we're going to have the function x2 minus f1, and f1 is just going to be the function evaluated at 1. So we're going to have f1=12, and that's simply equal to 1. Now all of this is going to be divided by x-1. Now 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 12-1 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. Well, what I can do is try factoring the top of this fraction, because x2-12 is the same thing as x2-1. One squared is just 1, so this is the same value. Now a difference of squares, I can factor to look like this. It would be (x+1)(x-1). I'm just using this rule for a difference of 2 squares, and all of this is going to be divided by x-1. Now from here, I can cancel the x-1s, so all I'm going to have is the limit as x approaches 1 for x+1. Now at this point, now 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. So 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 2 points on the graph rather than looking at 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 the 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

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

• A. -6
• B. 6
• C. 10
• D. -10

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

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

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 an 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. And in this video, we're going to now learn how to solve some other problems associated with tangent lines, specifically finding their equations. Now 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 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 a tangent line. Now 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 notice 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 squared 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 plus 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 in 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 \( 3 \times (x - 2) \). 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 minus 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 wanna calculate the slope, and then you wanna 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 \( y - 8 \). 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 \( x - \) 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 parentheses. So what we're going to end up with is -12 times x, which is \( -12x \), 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 \( -12x \), 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 $f\left(x\right)=3(x^2-1)$, find the equation of the tangent line at $x=1$.

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

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 \( f(x) = -4x^2 \) at \( x = -2 \). To start with this problem, what we need is 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 the tangent line.

Now I can go ahead and apply the function and the numbers that were given to this equation. I can see that \( c \) is going to be our given \( x \) value, which right there is \( -2 \), so we're going to have the limit as \( x \) approaches \( -2 \). This limit will stay consistent throughout this problem. We also have our function \( f(x) \), which is \( -4x^2 \) given to us. We have \( -4x^2 \) then minus \( f(c) \). Now \( f(c) \) is going to be, in this case, \( f(-2) \), and using this equation, \( f(-2) = -4(-2)^2 \). Evaluating \( (-2)^2 \) gives us \( 4 \), and \( -4 \times 4 = -16 \). So that means we're going to have \( -4x^2 - (-16) \) all divided by \( x - (-2) \). Now at this point, I can cancel these negative signs resulting in \( -4x^2 + 16 \) all divided by \( x + 2 \).

Now I can try applying my limit by substituting \( x = -2 \). One thing I notice here is that substituting \( x = -2 \) results in \( x + 2 = 0 \) on the bottom, and we cannot divide by zero in mathematics, so we're going to need a different strategy. I will try factoring the numerator. Pulling a \( -4 \) outside, we'll have \( -4(x^2 - 4) \) all divided by \( x + 2 \). Factoring \( x^2 - 4 \) as a difference of squares gives \( (x-2)(x+2) \), then the equation simplifies to \( -4(x-2)(x+2) \) over \( x+2 \). Now I can cancel the \( x+2 \) terms, leaving \( -4(x - 2) \).

Now, I can apply my limit since we've canceled the expression in the denominator. We have the slope of our tangent line as \( -4(-2 - 2) \). Calculating \( -2 - 2 \) gives \( -4 \), and \( -4 \times -4 = 16 \). So, the slope of our tangent line is \( 16 \).

Since the problem asks us to find the equation of the tangent line, we'll use the point-slope form of a line \( y - y_1 = m(x - x_1) \). We know \( m = 16 \) and the tangent point is \( (-2, -16) \). Substituting the values gives \( y - (-16) = 16(x - (-2)) \). Simplifying, we have \( y + 16 = 16(x + 2) \), distributing gives \( y + 16 = 16x + 32 \). Solving for \( y \), subtract 16 from both sides to get \( y = 16x + 16 \). This line equation, \( y = 16x + 16 \), is the equation of our tangent line.

That is the solution to this problem. Hope you found this video helpful.

Derivatives

So in recent videos, we got introduced to this idea of a derivative, and recall that a derivative is just the slope of a tangent line at a certain value, like x=1 or x=-2. Now, in this video, we're going to learn how we can find this derivative, AKA the slope of a tangent line, at any possible x value, at any possible point. And that might sound a bit scary, but it turns out this process for finding the general equation of a derivative is actually pretty similar to the process we took for finding the slope of a tangent line. So without further ado, let's just jump right into an example of what one of these problems might look like in this course.

So here we have this example where we're asked to find the derivative of this function, f(x)=x2, for any x value. And we're asked to use this to find the slope of the tangent line for these two specific values, x=1 and x=-2. Now our first step when solving these types of problems is to find the general equation because that's going to give us the slope of the tangent line for any possible x. Now to do this, what I need to do is use the equation for a derivative. And something that I'll mention here is that whenever you see derivatives, they could be written in this notation. These are all types of things you could see in your textbook. This little apostrophe here is what we call prime. So, I want to find the derivative of our function, it's going to be f′x. And the derivative of our function is going to be the limit as h approaches 0 for fx+h-fxh. So this is the equation you're going to need, and you may want to write this down.

Now a question that will come up about this formula right here is what exactly is this h that we're seeing? Right? Because this is not a variable we're really familiar with. Well, it turns out all this h is, is it's basically just taking the place of this x−c that we saw in the previous equation, where we had two points and they slowly converge to one specific value that we were looking at. Well, in this equation, since we're trying to find the slope of the tangent line for any possible x, this h here is any possible two points. And since we know they're going to get really close to converge at one value, we say this limit of h approaches 0. So let's go ahead and see if we can use this equation to find the general equation for a derivative.

So we have our function f(x)=x2. And if I want to use this equation up here, well, what I'm going to do is plug in these variables. So what I see is that our function is x2. So fx+h is going to be x+h2. And then we're going to have minus fx, which I can see is x2, all divided by h. So that right there is using this equation and applying it to this example. Now what I can try to do is evaluate this limit, but I noticed that we have h approaching 0. If I try to put 0 in right there, we're going to be dividing by 0, which is a math rule that we cannot break. So what I need to do here is I need to find some sort of way to get this h to go away in the denominator. And to do this, well, I'm first going to start by factoring this expression right here. So we have x+h2, and that can also be written as x2+2xh+h2, and then this is going to be minus this x2 we have here, all divided by h. So this portion right here is just me expanding x+h2 out. But notice when we do this, I can actually get rid of this x2 that I see. And I can also take this h and factor it out of each of these terms. So what I'm going to end up having is h×2x+h all divided by h. Now this is a great position to be in because notice I can now cancel one of the hs. So all we're going to have is 2x+h. Now at this point, I can apply our limit by taking 0 and plugging it in for h. That's going to give us 2x+0, which is simply equal to 2x. So this right here is the general equation for the derivative, and that would be the derivative at anypossible x value right there. That's the solution.

Now we're not done with this problem yet, because we're also asked to find what the slope of the tangent line is at x=1 and x=-2. Now to do this, well, we just need to use this result that we got right here. Because using this equation right here called the definition of the derivative, we were able to find the derivative which is the slope of the tangent line. So what I'm just going to do is take each of these values, x=1 and x=-2, and plug it into this equation. So if I find f′1, that's going to be evaluating where x equal to 1. So I just need to replace this x here with 1. So we're going to have 2×1, which is equal to 2. And that right there is the slope of the tangent line at x=1. Now notice something. When we solved this type of problem before, the slope of our tangent line at x=1 was also 2 for the function x2. But rather than having to go through this whole process of finding these two points, well, once we found the general derivative, we could just plug this number directly in, and it gave us the same result. But now let's try a different value, like x=-2. Well, to find this, I just need to take this x right here and replace it with -2. So we're gonna have 2×-2, which is equal to -4. So that right there is the slope of the tangent line at x=-2. And so if we were to go onto our graph here and we were to draw a tangent line right about here, this tangent line would have a slope of -4. And that is what this equation does. This general equation allows us to find the derivative at any possible x value. And the nice thing is, we don't even have to stop here. If we wanted to, we could figure out what the slope of the tangent line is at x=100, or x=1000, or even x=-56, or just something random. We could plug whatever x value we want into this equation right here, and it would give us the slope of the tangent line at that specific value. So that is how you can find the general equation for the derivative as well as the derivative at certain x values using this definition right here. So I hope you found this video helpful, and let's try getting some more practice.

Find the derivative of the function $f\left(x\right)=4x^2-9x$.

• A. $f^{\prime }\left(x\right)=4x-18$
• B. $f^{\prime}\left(x\right)=8x-9$
• C. $f^{\prime }\left(x\right)=0$
• D. $f^{\prime }\left(x\right)=4x-9$

Use the definition of a derivative, to find the derivative of the function $g(x)=x^3$ at $x=-1$.

• A. 0
• B. -1
• C. -3
• D. 3