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Differential Calculus (2017 edition)

Course: Differential Calculus (2017 edition)>Unit 5

Lesson 2: Derivative as slope of tangent line

The derivative & tangent line equations

Discover how the derivative of a function reveals the slope of the tangent line at any point on the graph. We'll explore how to use this powerful tool to determine the equation of the tangent line, enhancing our understanding of instantaneous rates of change.

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• I thought a tangent line could only touch one point on the graph?
• Technically, a tangent line is one that touches a curve at a point without crossing over it. Essentially, its slope matches the slope of the curve at the point. It does not mean that it touches the graph at only one point. It is, in fact, very easy to come up with tangent lines to various curves that intersect the curve at other points.

The key here is that the tangent should only touch the curve at the point of interest.
• `g'(-1)=-2` -- doesn't that mean that the tangent goes through (-1,-2)? Sal's tangent doesn't.
• I was confused by this as well. I was thinking, incorrectly, that "g'(-1) = -2" means that the tangent line must go through (-1, -2), as you said. But the derivative, g', doesn't tell us what points the tangent line goes through. It tells us only the slope of the line for a given x. The value of the slope of the tangent line could be 50 billion, but that doesn't mean that the tangent line goes through 50 billion. In fact, the tangent line must go through the point in the original function, or else it wouldn't be a tangent line. The derivative function, g', does go through (-1, -2), but the tangent line does not. It might help to think of the derivative function as being on a second graph, and on the second graph we have (-1, -2) that describes the tangent line on the first graph: at x = -1 in the first graph, the slope is -2.
• I might be missing something, but why did we choose to represent the line with equation y=mx+b ?
Could we use any other equation and still get the same result for value of 'b' ?
• He chose to use y=mx+b because a tangent line, or the derivative of a function will always be a straight line, and that equation (y=mx+b) is how we show the line. The 'b' value is just the y-intercept. It is where the line hits the y/vertical axis. But I'm just starting to study this stuff so don't rely on my answer too much lol. Good luck.
• I ran into this question on a quiz:

The tangent line to the graph of function g at the point (-6, -2), passes through the point (0,2).
Find g'(-6).

The correct solution is:
((-2)-2) / (-6-0) = -4 / -6 = 2/3

But could it also be this?
(2-(-2)) / 0-(-6) = 0 / -6 = 0

Thanks!
• Your setup in the second version is correct, but 2-(-2)=4, and 0-(-6)=6, so the slope is still 4/6=2/3.
• I'm happy with the y = mx + c form, but in the questions that follow this section, they were of the form (y + a) = b(x + d).
It wasn't obvious to me how to convert one to the other.
• That seems to be the point-slope form. It's used to get the equation when a point and the slope are given. This is actually derived from the slope formula itself.

Let the slope = m and let point (p,q) exist on the line. We need to find the equation of the line. Let's take an arbitrary point (x,y). So, from the definition of a slope.

m = (y-q)/(x-p)

Rearranging, you get (y-q) = b(x-p)

And that's exactly what you wrote, just with different variables.

Honestly speaking, such questions can almost always be done with the definition of a slope. So, you don't really need to bother learning all the different formulae. They're just iterations of the same thing.
• consider a function f(x)=x^2 its derivative will be f'(x)=2x and f'(2) will be 4
am i right?
• Couldn't you have also found the y-intercept of the tangent line just by looking at, y'know, where it intercepts the y-axis? I knew from looking at the line that b = 1 because it's very obvious that the line intercepted the y-axis at y = 1. Did he do the algebra just for the sake of clarification or have I been finding y-intercepts wrong my whole life? lol
• A lot of times it won't be so easy to tell where the y intercept is. Basically when the y intercept is not exactly on a marked number.
• Isn't "b" in y=mx+b where the function crosses the Y-axis?
• Yes it is. The tangent line is the straight green line he drew, which infact crosses the y-axis at y=1
• please solve this for me and explain it??
Find the value of a, b, c for which the parabola ax^2+bx+c passes through (1, 1) and has a tangent with gradient 7, at point (3, 3).
y- f(a) = f'(a)(x-a)
• Plugging in your point (1, 1) tells us that a+b+c=1.
You also say it touches the point (3, 3), which tells us 9a+3b+c=3. Subtract the first from the second to obtain 8a+2b=2, or 4a+b=1.

The derivative of your parabola is 2ax+b. When x=3, this expression is 7, since the derivative gives the slope of the tangent. So 6a+b=7.

So we have
6a+b=7
4a+b=1

Subtract the second equation from the first to get 2a=6, or a=3. Substitute back in to get 12+b=1, or b= -11.

Since a+b+c=1, this gives us 3-11+c=1
-8+c=1
c=9