Hi guys! I'm Nancy..
and I'm going to show you how to find the equation of a tangent line using derivatives.
What is a tangent line? Well, it just means if you have a graph...
with some curve on it
the tangent line is the line
that touches your graph at one point exactly.
It's a linear line
that is tangent to your curve and sort of grazes it and touches at one point.
So tangent line problems can come in many forms
but thankfully you can use the same four steps in all of them
so let me show you some examples.
OK. Say you have a problem that says to find
the tangent line to Y equals X cubed plus 2X at a point
at the point: 1, 3. How do you start this problem? Well
it's always gonna be the same four steps in these tangent line problems.
The first step is to take the derivative of the equation that you're given.
This Y equals
is also F of X.
It's also a function.
You want to take the derivative of the equation you're given, so F prime of X is your first step.
F prime of X
and then you take the derivative of this expression.
The derivative of X cubed...
remember for the power rule, you bring the power down and then decrease that power by 1.
so for the first term, we have 3X squared
for the second term, the derivative of 2X is just 2. The X goes away.
You have 3X squared plus 2. So that's your derivative, your F prime of X.
The second step is always to take the X coordinate that you're given in the point...
and plug it in to F prime of X.
You can call that F prime of 1.
Since you're X coordinate is 1. So you take that number, and you're going to plug it in
to what you just found, the derivative. And this is just notation for that, for plugging in 1.
Anyway, when you plug in 1 for X, you have 3 times 1 squared plus 2...
which simplifies to 3 times 1 plus 2
which is just 5, so F prime of 1...
equals 5.
This is going to be your slope number that you'll use.
The third step is...
to find the Y coordinate of your point, if you're not already given it
so sometimes you're not given the Y.
In this problem, you're given the whole point, which could happen.
So you have Y equals 3, so you don't need to find Y in this case.
You can move on to the fourth step, the last step, is to plug in the numbers that you've found
into the point-slope formula for a line.
The point-slope formula...
in general looks like Y minus Y1 equals M times X minus X1.
M is just your slope
and that's going to be the F prime value that you found when you plugged in X
so your M in this problem is 5.
Which you found right here.
X1 and Y1 are the coordinates of the point that you found or that you were given.
So this here you can also call X1, Y1.
You have both of those numbers.
Now you want an equation for the tangent line.
That means that in your equation you need to leave an X variable and Y variable in it.
So it's OK to leave those there. In fact you want it.
You can go ahead and you can write your equation
by plugging in values for M, for X1, and for Y1.
So your X1 is the coordinate 1.
Your Y1 is the Y coordinate of the point.
The original point, which is 3.
So if you rewrite this, your equation is
Y - 3 equals the slope 5, times (X - 1).
Notice you're leaving X and Y in the equation
because you want it to be an equation for Y in terms of X.
So this... this is correct. This is a correct answer.
Sometimes it's considered simpler to move it around
and change it to be in slope intercept form
or Y = MX + B form.
So you can do that with some simple algebra
if you were to multiply, distribute out, the... 5X - 5, and rearrange the numbers
you could get into Y equals form.
if you were to multiply, distribute out, the... 5X - 5, and rearrange the numbers
you could get into Y equals form.
Y = 5X - 5 + 3... which is just minus 2.
So this.. Y = 5X - 2 is your equation of the tangent line to this F of X at that point.
Alright, say you have F(X) = X / (X - 1)
and you have to find the equation of the tangent line at X = 2.
This looks a lot more complicated
but the truth is it's the same four steps
no matter how complicated your F(X), your equation gets.
but the truth is it's the same four steps
no matter how complicated your F(X), your equation gets.
This could be ridiculous..
but as long as you know how to take F prime of X, the derivative, you can still
do all the same steps and get the equation of the linear tangent line.
So the first step is still, find F of X.
Find the derivative of your given equation.
You want F prime of X..
Now since your equation is X / (X - 1)
this is a fraction. This is a quotient.
so you'll have to use the quotient rule to take the derivative.
Which is a little more complicated than you might like..
but the steps are pretty straightforward.
Think back to doing the quotient rule..
It's the bottom of the function times the derivative of the top..
derivative of X is just 1..
..minus the top expression, X, as is..
times the derivative of the bottom, which is 1 from the X
and nothing from the -1, so it's just 1 there.
And then it's all over the bottom squared.
The lower function squared. (X - 1) squared.
OK. So this is the derivative, but you should simplify it
because things will cancel.
Try simplifying within the top first. Within the numerator.
This is X - 1 - X ... another X...
all over (X - 1) ^ 2.
Great news is that something cancels on top. This becomes simpler.
X - X cancels and you're just left with -1 on top.
So, the simplest way to write your derivative F prime of X is..
-1 / (X - 1)^2
Remember F prime just means the derivative.
Alright. Second step.
Again, plug in your X value, into the derivative F prime of X.
What you can call that is F prime of 2, because you're plugging in 2.
So the notation for that is..
F prime of 2 means you're putting 2 in place of every X in your derivative expression.
So in this case, that means -1 on top...
and (2 - 1)^2 on the bottom.
This simplifies to...
-1 over 1^2, which is just 1.
So your slope value...
your M is -1.
I'll make that clear.
OK. Third step. Find the original Y value if you don't already have it.. if you weren't given it.
You weren't a full point, you were only half, just the X coordinate.
So you have to find Y. How do you do that?
All you need to is take the original X, plug it in to your original equation, your original F...
and find Y, find the Y value.
So you can call that F of 2
That will give you your Y. F of 2
means you're plugging in 2 for X, so you have 2 / (2 - 1)...
which is 2 / 1, which is 2.
Alright, so that is your Y.
So if you had to write a full point for the original point...
it would be 2 for X and 2 for Y, or (2,2).
That's important, you're going to use both of those numbers next.
Because the last step is still to plug in your M, your Y, and your X...
all into the point-slope form for the equation of a line.
And that point-slope form looks like...
Y minus your Y coordinate, your Y1, which is 2...
equals M, your slope, which is -1...
times parentheses X, the X you want that in the equation. Otherwise it's not a full equation...
minus the X coordinate, or X1, which is also 2 in this problem.
Alright, so you know, this is technically correct, you can leave your answer in point-slope form.
Unless you have to rearrange it to get it into Y = MX + B form, but technically this is correct
so this is the equation for the tangent line to this graph at X = 2.
OK. One final problem.
I'm giving you one that has a trig function in it.
F of X equals (sec X)^2
There's also a power in here, so this might look kind of scary, maybe...
but just rely on what you know about derivatives, differentiating.
In this case you're going to have to use the Chain Rule..
..yay chain rule..
because there's a function inside of a power.
but even if this is a more complicated derivative
you're still going to do the same steps. The first step is still...
find F prime of X. Find the derivative.
F prime of X
Like I said, this is a Power Rule with another function inside, so...
use the Chain Rule because you don't just have X inside. You have some other X expression
inside a power, you have to use the Chain Rule.
First, you take the derivative of the power using the Power Rule.
So the 2 power comes down out front as a coefficient, 2.
You can still put parentheses. Sec X is still there...
this power 2 is reduced by 1, so the power is 1. We don't need to write that.
If we don't write it, it's assumed that the power is 1.
You're not done. You are not done, because it's not just X inside.
Since there's a sec X, you also need to multiply by the derivative of sec X..
the inside function.
You did the derivative of the outside function. Now you need to multiply by the derivative of the inside function.
What is the derivative of sec X?
Well you can get that from a derivative table
with the trig derivatives, common trig derivatives, or maybe you just know it.
The derivative of sec X is...
Sec X...
times Tan X
Sec X Tan X
This is your F prime of X.
Alright, second step.
Plug in the X value that you're given. You're given pi over 3 for your X value.
Plug it in to F prime.
So you have F prime of pi over 3.
It's a weird pi number for your X value..
but that's totally valid, especially for a trig function
to have a pi/3X value that you're plugging in.
so this would be 2 times sec(pi/3)...
times another sec(pi/3)...
tangent...
of pi/3.
Alright, this can be simplified.
This should come out to be just a number, a constant
so you don't want to leave it like this.
As for secant...
It's always easier to find cosine first and then take the reciprocal, flip it.
It's easier to evaluate secant if you first do cosine
because secant is 1 / cos, so we have 1 / cos(pi/3)...
times another 1 / cos(pi/3)...
then you can just leave tan(pi/3).
Unless you prefer to write sin over cos.
But leave it as tan(pi/3).
Now you need to use what you know about the unit circle...
which you may remember
cos(pi/3)... pi/3 is up here.
cos(pi/3) is the x coordinate of pi/3
which is one half.
So this becomes 2 times...
1 over (1/2)...
times 1 over (1/2)...
times tan(pi/3). The tangent of pi/3...
is root 3.
So it's times square root of 3.
These values - one half, one half, root 3 - you can also get them from your calculator
if you're allowed to use a calculator. Just make sure that you're in radian mode
before you plug in cos(pi/3).
This simplifies because you have 2...
1 over (1/2)
Dividing by a fraction is the same as multiplying by its reciprocal
so it's the same as multiplying by 2.
So this is 2, and this is 2.
So that's 4.
2... times 2. Four.
And then you have a root 3 left over.
So your final F prime of pi/3, slope, value is
8 root 3. You can combine the 2 times 4
And that right there is your M, slope, value.
Alright, that was step two.
Step three, remember, you probably already have figured this out..
step three is to find the Y value, original Y value
if you're not given it.
But here you were given the full point, so Y is 4.
Your Y1 in the point-slope form is 4.
You have your X1...
and you have your M, so now you're ready to plug into the point-slope form
as your final step.
And that is Y minus Y1, which is 4
equals M, which is 8 root 3, you can just write that.
You don't want to write the decimal version of that.
8 root 3 times X, you want X in your equation
minus X1, which is pi/3.
You can just write that, pi/3.
And this right here is a totally valid way of writing your final answer.
It's in point-slope form.
Rearrange it if you have to, but this is a valid answer for
the equation of the tangent line to this F of X, at that point.
So I hope this video helped you
figure out how to write the equation of a tangent line using derivatives.
Look.. anything that makes calculus easier is a good thing.
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