Calculating slope of tangent line using derivative definition | Differential Calculus | Khan Academy
Summary
TLDRThis video script explains the concept of finding the slope of a curve at a specific point by using the slope of a secant line between two points. As the distance between the points decreases, the secant line's slope approaches the slope of the tangent line. The derivative of a function, represented as f'(x), is introduced as the slope of the tangent line. The example uses the function y = x^2 to find the slope at x = 3, demonstrating the process of taking the limit as the secant line approaches the tangent line, resulting in a slope of 6.
Takeaways
- 📐 The concept of slope of a curve at a specific point is introduced.
- 🔍 The slope of a secant line is calculated by taking two points on the curve, with one point not too far from the other.
- 📈 The change in y (Δy) divided by the change in x (Δx) gives the slope of the secant line.
- 🎯 The slope of the tangent line at a specific point is approached as the secant line gets closer to that point.
- 🧭 The derivative of a function at a point is defined as the slope of the tangent line at that point.
- 📉 The process of finding the slope at a specific point is demonstrated using the curve y = x^2 at x = 3.
- 🔢 The notation 'delta x' (Δx) is used to represent a small change in x, which is a common notation in calculus.
- 📋 The slope of the secant line simplifies to 6 + Δx when considering the curve y = x^2.
- 📉 As Δx approaches 0, the slope of the secant line approaches the slope of the tangent line, which is the derivative of the function at that point.
- 🔄 The limit as Δx approaches 0 of the slope of the secant line gives the slope of the tangent line, which is 6 at x = 3 for y = x^2.
Q & A
What is the main concept discussed in the video script?
-The main concept discussed in the video script is the calculation of the slope of a curve at a specific point, which is the derivative of a function at that point.
How is the slope of a curve at a point determined?
-The slope of a curve at a point is determined by finding the slope of the secant line between that point and another point close to it, and then taking the limit as the second point approaches the first point.
What is the secant line in the context of this video?
-The secant line in this video is a line that passes through two points on the curve, and its slope is calculated using the change in y divided by the change in x between these two points.
How is the slope of the secant line related to the slope of the tangent line?
-The slope of the secant line approaches the slope of the tangent line as the second point on the curve gets closer to the first point.
What does the derivative of a function represent?
-The derivative of a function represents the slope of the tangent line to the curve of the function at a specific point.
What is the function used as an example in the video script?
-The function used as an example in the video script is f(x) = x^2.
At which point on the curve does the video script focus on finding the slope?
-The video script focuses on finding the slope of the curve at the point where x equals 3.
What is the significance of the notation change from 'h' to 'delta x'?
-The notation change from 'h' to 'delta x' is to expose viewers to different notations used in various books and to emphasize that the concept remains the same regardless of the symbol used.
What is the formula for the slope of the secant line in terms of delta x?
-The formula for the slope of the secant line in terms of delta x is (f(x + delta x) - f(x)) / ((x + delta x) - x), which simplifies to (9 + 6 delta x + delta x^2 - 9) / delta x.
How does the slope of the secant line simplify when delta x approaches 0?
-When delta x approaches 0, the slope of the secant line simplifies to 6, which is the derivative of the function f(x) = x^2 at x = 3.
What is the limit process described in the video script?
-The limit process described in the video script is taking the limit of the slope of the secant line as delta x approaches 0, which results in the slope of the tangent line.
Outlines
📐 Calculating the Slope of a Tangent Line
This paragraph explains the process of finding the slope of a curve at a specific point by using the concept of a secant line. The presenter starts by discussing the slope between two points on a curve, where one point is kept fixed and the other moves closer to it. As the distance between the points decreases, the slope of the secant line approaches the slope of the tangent line at that point. The concept of the derivative, denoted as f'(x), is introduced as the slope of the tangent line. The presenter then illustrates this with an example using the function y = x^2 and the point where x = 3. By selecting another point on the curve at x = 3 + delta x, the slope of the secant line is calculated as the change in y over the change in x. The change in y is expressed as the difference between the y-values of the two points, which simplifies to 6 + delta x when divided by delta x.
🔍 Simplifying the Secant Line Slope to Find the Tangent
The paragraph continues the discussion on finding the slope of the tangent line at a specific point on the curve y = x^2. The presenter simplifies the expression for the slope of the secant line by dividing both the numerator and the denominator by delta x, resulting in the slope being 6 + delta x. This expression is then used to explore what happens as delta x approaches zero. The presenter explains that as delta x gets smaller, the secant line more closely approximates the tangent line. By taking the limit as delta x approaches zero, the slope of the tangent line at x = 3 is found to be 6. This is confirmed by setting delta x to zero in the expression for the secant line slope. The presenter concludes by stating that they will provide a general formula for the slope of the tangent line at any point on the curve in the next video.
Mindmap
Keywords
💡slope
💡secant line
💡tangent line
💡derivative
💡limit
💡change in y
💡change in x
💡curve
💡function
💡delta x
💡foiling
Highlights
Introduction to finding the slope of a curve at a specific point using the secant line method.
Explanation of the secant line's slope as the change in y divided by the change in x.
The concept of taking the limit as the secant line approaches the tangent line to find the derivative.
Derivative defined as the slope of the tangent line at a specific point on a curve.
Application of the derivative concept to a concrete example with the function y = x^2.
Determining the slope of the tangent line at the point where x = 3 for the curve y = x^2.
Use of the variable 'delta x' to represent a change in x values for calculating the slope.
Calculation of the y-value for a point on the curve y = x^2 given an x value.
Derivation of the slope formula for the secant line between two points on the curve.
Simplification of the secant line's slope formula to isolate the variable 'delta x'.
Explanation of how the slope of the secant line approaches the slope of the tangent line as 'delta x' approaches zero.
Limit process to find the exact slope of the tangent line at a specific point.
Result of the limit process showing the slope of the tangent line at x = 3 is 6.
General formula for the slope of the tangent line for the function y = x^2 at any point x.
Introduction to the concept of the derivative as f'(x) for the function y = x^2.
Anticipation of a general formula for the slope of the tangent line at any point on the curve in the next video.
Transcripts
In the last video we tried to figure out the slope of a
point or the slope of a curve at a certain point.
And the way we did, we said OK, well let's find the slope
between that point and then another point that's not too
far away from that point.
And we got the slope of the secant
line.
And it looks all fancy, but this is just the y value of the
point that's not too far away, and this is just the y value
point of the point in question, so this is
just your change in y.
And then you divide that by your change in x.
So in the example we did, h was the difference
between our 2 x values.
This distance was h.
And that gave us the slope of that line.
We said hey, what if we take the limit as this point
right here gets closer and closer to this point.
If this point essentially almost becomes this point, then
our slope is going to be the slope of our tangent line.
And we define that as the derivative of our function.
We said that's equal to f prime of x.
So let's if we can apply this in this video to maybe make
things a little bit more concrete in your head.
So let me do one.
First I'll do a particular case where I want to find the
slope at exactly some point.
So let me draw my axes again.
Let's draw some axes right there.
Let's say I have the curve-- this is the curve-- y
is equal to x squared.
So this is my y-axis, this is my x-axis, and I want to know
the slope at the point x is equal to 3.
When I say the slope you can imagine a tangent line here.
You can imagine a tangent line that goes just like that, and
it would just barely graze the curve at that point.
But what is the slope of that tangent line?
What is the slope of that tangent line which is the same
as the slope of the curve right at that point.
So to do it, I'm actually going to do this exact technique that
we did before, then we'll generalize it so you don't have
to do it every time for a particular number.
So let's take some other point here.
Let's call this 3 plus delta x.
I'm changing the notation because in some books you'll
see an h, some books you'll see a delta x, doesn't hurt to
be exposed to both of them.
So this is 3 plus delta x.
So first of all what is this point right here?
This is a curve y is equal to x squared, so f of x is 3
squared-- this is the point 9.
This is the point 3,9 right here.
And what is this point right here?
So if we were go all the way up here, what is that point?
Well here our x is 3 plus delta x.
It's the same thing as this one right here,
as x naught
plus h.
I could have called this 3 plus h just as easily.
So it's 3 plus delta x up there.
So what's the y value going to be?
Well whatever x value is, it's on the curve, it's going
to be that squared.
So it's going to be the point 3 plus delta x squared.
So let's figure out the slope of this secant
line.
And let me zoom in a little bit, because that might help.
So if I zoom in on just this part of the curve,
it might look like that.
And then I have one point here, and then I have the
other point is up here.
That's the secant
line.
Just like that.
This was the point over here, the point 3,9.
And then this point up here is the point 3 plus delta x, so
just some larger number than 3, and then it's going to
be that number squared.
So it's going to be 3 plus delta x squared.
What is that?
That's going to be 9.
I'm just foiling this out, or you do the distribute
property twice.
a plus b squared is a squared plus 2 a b plus b squared, so
it's going to be 9 plus two times the product
of these things.
So plus 6 delta x, and then plus delta x squared.
That's the coordinate of the second line.
This looks complicated, but I just took this x value and I
squared it, because it's on the line y is equal to x squared.
So the slope of the secant
line is going to be the change in y divided
by the change in x.
So the change in y is just going to be this guy's y value,
which is 9 plus 6 delta x plus delta x squared.
That's this guy's y value, minus this guy's y value.
So minus 9.
That's your change in y.
And you want to divide that by your change in x.
Well what is your change in x?
This is actually going to be pretty convenient.
This larger x value-- we started with this point on the
top, so we have to start with this point on the bottom.
So it's going to be 3 plus delta x.
And then what's this x value?
What is minus 3?
That's his x value.
So what does this simplify to?
The numerator-- this 9 and that 9 cancel out,
we get a 9 minus 9.
And in the denominator what happens?
This 3 and minus 3 cancel out.
So the change in x actually end up becoming this delta x, which
makes sense, because this delta x is essentially how much more
this guy is then that guy.
So that should be the change in x, delta x.
So the slope of my secant
line has simplified to 6 times my change in x, plus my change
in x squared, all of that over my change in x.
And now we can simplify this even more.
Let's divide the numerator and the denominator
by our change in x.
And I'll switch colors just to ease the monotony.
So my slope of my tangent of my secant
line-- the one that goes through both of these-- is
going to be equal if you divide the numerator and
denominator this becomes 6.
I'm just dividing numerator and denominator by delta
x plus six plus delta x.
So that is the slope of this secant
line So slope is equal to 6 plus delta x.
That's this one right here.
That's this reddish line that I've drawn right there.
So this number right here, if the delta x was one, if
these were the points 3 and 4, then my slope would be 6 plus
1, because I'm picking a point 4 where the delta x here
would have to be 1.
So the slope would be 7.
So we have a general formula for no matter what my delta
x is, I can find the slope between 3 and 3 plus delta x.
Between those two points.
Now we wanted to find the slope at exactly that
point right there.
So let's see what happens when delta x get
smaller and smaller.
This is what delta x is right now.
It's this distance.
But if delta x got a little bit smaller, then the secant
line would look like that.
Got even smaller, the secant
line would look like that, it gets even smaller.
Then we're getting pretty close to the slope
of the tangent line.
The tangent line is this thing right here that I
want to find the slope of.
Let's find a limit as our delta x approaches 0.
So the limit as delta x approaches 0 of our
slope of the secant
line of 6 plus delta x is equal to what?
This is pretty straightforward.
You can just set this equal to 0 and it's equal to 6.
So the slope of our tangent line at the point x is equal to
3 right there is equal to 6.
And another way we could write this if we wrote that f of
x is equal to x squared.
We now know that the derivative or the slope of the tangent
line of this function at the point 3-- I just only evaluated
it at the point 3 right there-- that that is equal to 6.
I haven't yet come up with a general formula for the slope
of this line at any point, and I'm going to do that
in the next video.
Посмотреть больше похожих видео
Average vs Instantaneous Rates of Change
Derivative as a concept | Derivatives introduction | AP Calculus AB | Khan Academy
Applying First Principles to x² (2 of 2: What do we discover?)
Lesson 4-1, Video 5; Perpendicular Line 2
Graphing a line given point and slope | Linear equations & graphs | Algebra I | Khan Academy
Introduction to point-slope form | Algebra I | Khan Academy
5.0 / 5 (0 votes)