If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

### Course: AP®︎/College Calculus AB>Unit 2

Lesson 7: Derivative rules: constant, sum, difference, and constant multiple: connecting with the power rule

# Tangents of polynomials

Sal finds the equation of the line tangent to the graph of f(x)=x³-6x²+x-5 at x=1.

## Want to join the conversation?

• The function here is cubic. The derivative is quadratic. I don't understand why evaluating f'(1) gets us the slope of the tangent line at 1? F' in this case isn't a line?!
(9 votes)
• Derivatives don't have to be linear to still give us the slope of the tangent line. The point is that the derivative is a function that returns a single value at any point, which represents the slope of the tangent. The reason this works is shown in the proof videos - i.e., the ones showing the derivative expressed as the limit of a secant slope.

Consider this: if the slope of a function `f'` is changing in a way that is non-linear, how could you expect to find a linear function `f'` that tells you the slope of `f` at any point? By definition, that's not possible. So the derivative of a function whose slope changes nonlinearly will itself be nonlinear (by definition).

I hope that makes sense and somewhat answers your question.
(38 votes)
• At the constant x's power one meant the derivative was one, but I remember in one of the first videos covered that the derivative of any constant is zero. What made it one here?
(5 votes)
• The derivative of a constant is 0, but x is not a constant. It's a variable. So we apply power rule, which works for functions of the form x^n.
(10 votes)
• @　Why does the derivative of a constant become ZERO?

Is it becaue the slope of y = -5 is zero?
(5 votes)
• Yes, that is the idea. More generally the derivative of any constant c is 0, because the slope of y = c is zero.

Have a blessed, wonderful day!
(5 votes)
• How would you find the derivative if there are fractions? For example: 3/x^2 -4/x^3?
I appreciate for the help.
(3 votes)
• For the ones you gave, it's important to remember that 1/(x^a) = x^(-1). For the the derivative of 3/(x^2), you'd treat it as 3x^(-2) and use the power rule to get -6x^(-3). The rule also applies for fractional exponents like x^(3/2); it's derivative is (3/2)x^(1/2).
(5 votes)
• What exactly is the derivative? How does it mathematically relate to solving the tangent. Also, do we use the derivative to solve anything else?
(2 votes)
• The derivative is the rate of change so dy/dx is an infinitesimal change of y over an infinitesimal change in x. I would suggest you take look at differentiation by first principles to fully understand it.

So let say there is a car and you have want instantaneous speed of the car at 2.5 seconds. To find the speed of car we need to find the rate of change based on x and t where t is time and x is the distance travelled. So the average speed capital delta x / capital delta t s in physics as the capital delta is just difference in the distance and time so it (x_1 - x)/ (t_1-t). However we need to find the instantaneous speed. However the speed will be the same if we find the speed right after 2.5 seconds as in find the speed at 2.5^+ seconds so that infinitesmal difference in the times. Finding the average speed then would give the exact speed if you were try find the speed between the two points

So we can say the instantaneous speed of the car when t=2.5 seconds is

lim_dt->0 ( f(2.5+ dt) - f(2.5) ) / (2.5+dt -2.5) =

(f(2.5 + dt) -f(2.5)) / dt
(4 votes)
• Why didn't Sal just use point-slope form there at the end? You find a point and a slope (or derivative), so it seems much more intuitive to me to do it that way instead of plugging the pieces into the form y = mx +b. I guess you could do that. But on the AP exam, point-slope form is a perfectly acceptable form and there might even be multiple choice questions in that format.
(2 votes)
• It doesn't matter which form is used as both will describe the same line. Personally, I prefer slope-intercept form as it seems far mare intuitive to me than point-slope form in any case.
(4 votes)
• For finding values of x where the tangent line to the graph is horizontal using the equation 2x^2 + 3x - 1 would you simply find the derivative of the equation and set that answer to 0? And if so, the answer would be -3/4?
(2 votes)
• That's correct. The statement "the derivative of 2x^2+3X-1 is 0" implies "x= -3/4".
(3 votes)
• When Sal took the derivative, he ended up with a quadratic polynomial. Why can we just plug in a number in that quadratic polynomial and come up with a slope? I wish he went into more detail that. Thanks
(2 votes)
• Yes, you can just input the number but Sal just wanted to show another way to solve this question
(2 votes)
• where does (y= mx+b) come from?
(2 votes)
• How does the graph of the derivative of f(x) look? And why?
(2 votes)

## Video transcript

- [Voiceover] What you see here in blue this is the graph of Y is equal to F of X. Where F of X is equal to X to the third minus six X squared plus X minus five. What I want to do in this video is think about what is the equation of the tangent line when X is equal to one? So we can visualize that. So, this is X equaling one right over here. This is the value of the function. When X is equal to one. Right over there. And then the tangent line looks something like will look something like. I know I can do a better job than that. It's going to look something like that. And what we want to do is find the equation the equation of that line. And if you are inspired I encourage you to be, pause the video and try to work it out. Well the way that we can do this is if we find the derivative at X equals one the derivative is the slope of the tangent line. And so we'll know the slope of the tangent line. And we know that it contains that point and then we can use that to find the equation of the tangent line. So let's actually just, let's just. So we want the equation of the tangent line when X is equal to one. So let's just first of all evaluate F of one. So F of one is equal to one to the third power which is one. Minus six times ones squared, so it's just minus six. And then plus one plus one minus five. So, this is equal to what? Two minus 11? Which is equal to negative nine. And that looks about right. That looks like about negative nine right over there. The scales are different on the Y and the X axis. So that is F of one. It is negative nine. Did I do that right? This is negative five. Yep, negative nine. And now let's evaluate what the derivative is at one. So, what is F prime of X? F prime of X. Well here it's just a polynomial. You take the derivative of X to the third while we apply the power rule. We bring the three out front. So you get three X to the. And then we go one less than three to get the second power. Then you have minus six X squared. So you bring the two times the six to get 12. So minus 12 X to the well two minus one is one power so that's the same thing as 12 X. And then plus the derivative of X is just one. That's just going to be one. And if you view this as X to the first power we're just bringing the one out front and decrementing the one. So it's one times X to the zero power which is just one. And then the derivative of a constant here is just going to be zero. So this is our derivative of F and if we want to evaluate it at one F prime of one is going to be three times one squared which is just three minus 12 times one which is just minus 12. And then we have plus one. So this is three minus 12 is negative nine plus one is equal to negative eight. So we know the slope right over here is the slope of negative eight. We know a point on that line it contains the point one negative nine so we can use that information to find the equation of the line. The line, just to remind ourselves, has the four. Y is equal to M X plus B. Where M is the slope. So we know that Y is going to be equal to negative eight X plus B. And now we can substitute the X and Y value that we know sits on that line to solve for B. So we know that Y is equal to negative nine. Let me just write this here. Y is equal to negative nine when X is equal to when X is equal to one. And so we get we get negative nine is equal to negative eight times one. So negative eight plus B. Well, let's see. We could add we could eight to both sides and we get negative one is equal to B. So we're done. The equation of the line the equation of this line that we have in magenta right over there that is that is Y is equal to the slope is negative eight X. And then the Y-intercept minus one.