Differentiability and Vertical Tangent Lines

Differentiability and Vertical Tangent Lines 1
Hi, everyone!
Welcome back to integralcalc.com.
Today we're going to be talking about differentiability of a function and how to find the horizontal and vertical tangent lines.
So in this particular problem, we've been asked to find the horizontal and vertical tangent lines
of this function f of x equals x times the square root of one minus x squared.
And after we find the horizontal and vertical tangent lines,
we’ve been asked to make a statement about the differentiability of this function.
In other words, where it's differentiable and where it isn't. So they've defined the range for us here.
We only care about this function between negative one and one.
So x greater than or equal to negative one. And less than or equal to positive one.
So I’ve gone ahead and graphed the function over here on the right.
You really don't need to graph it in order to solve anything about the problem but I thought it would be helpful for us visually.
So the way we're going to go about finding vertical and horizontal tangent lines is first by taking the derivative of our original function f of x.
So the derivative will be f prime of x
And in order to take the derivative, we will need product rule, which I’ve written over here.
Remember that product rule, we use whenever we've got two functions multiplied together, f of x and g of x.
In our case, we have f of x equals to x and g of x equal to the square root of one minus x squared.
We’re going to be using the righthand side of this formula here to take the derivative.
So notice that the first thing that we need is f prime of x, right?
So the derivative of f of x. Since f of x is equal to x here, the derivative is just going to be one.
And then we need the g of x function as it is without doing anything to it.
So that's going to be the square root of one minus x squared.
And then according to product rule, we add to that, f of x, which is just x, leave that alone.
And we multiply by the derivative of g of x. So let's pretend here for a moment. I'll write this over here.
We’ve got the square root of one minus x squared. That’s g of x, right?
Well, that’s the same thing as one minus x squared to the one-half, right?
That’s what the square root sign really means.
Whatever’s inside is raised to the one-half power.
So those two things are the same. So taking the derivative of that, we’ll need chain rule.
And remember that with chain rule, you take the derivative of the outside function first,
leaving the inside function completely untouched and then multiply by the derivative of the inside.
So taking the derivative of the outside, we’ll bring this exponent out in front here
so we’ll get one half and then we leave the inside function completely alone.
The inside function is one minus x squared. So we’re going to leave that alone and subtract one from our exponent.
So one-half minus one gives us a negative one-half. But now we need to multiply by the derivative of the inside.
Well, the derivative of one minus x squared is a negative two x. So we have to multiply here by negative two x.
And that’s how we take the derivative of the square root of one minus x squared.
So now, we can go ahead and simplify this.
This one is obviously redundant. We don’t need it.
We can just say the square root of one minus x squared.
And notice here, we’ve got a two in our denominator and a two in our numerator that are going to cancel.
We’ll just be left with this negative sign here. So instead of plus, we’ll say minus, to account for this negative sign.
We’ve got an x here and an x here that are going to become x squared so minus x squared.
And then we’re multiplying by one minus x squared to the negative one-half.
Well, we can move that to the denominator of our fraction and make it a positive one-half.
And then when we have a positive one-half, we know that it’s just equal to the square root so we get one minus x squared.
We get the square root of that in the denominator. That’s what this whole thing here converts to. Down here.
So, now we can go ahead…
Let’s find a common denominator so that we can combine this into one fraction.
We’ll multiply this first term over here by square root of one minus x squared,
and then divided by the square root of one minus x squared.
Because that will give us a common denominator here with our second fraction.
So when we do that, we’ll get the square root of one minus x squared times the square root of one minus x squared
is one minus x squared, over square root of one minus x squared and then we have minus x squared.
I’m just going to move this into the numerator here. x squared over the square root of one minus x squared.
Now, we can easily combine these fractions because we have a common denominator.
In the numerator, we’ll get one minus x squared minus x squared,
which is going to be one minus two x squared in the numerator all over the square root of one minus x squared.
Alright, so that’s our derivative, f prime of x.
now, to find horizontal tangent lines, we need to set the derivative equal to zero
and that’s because, remember that when you have a horizontal tangent line,
the slope of that tangent line or any horizontal line for that matter is going to be equal to zero.
A horizontal line has a slope of zero. So when the tangent line is horizontal, its slope is going to be zero.
And the derivative represents the slope of the tangent line.
So if we just set this equal to zero, we’ll find out where the slope of the tangent line is equal to zero.
So the only way that this equation here is going to be true, that this left hand side is going to be equal to zero,
is if the numerator of our fraction is equal to zero. If we get that equal to zero, the whole fraction is going to be zero.
So we can just say one minus two x squared equal to zero.
We’ll add two x squared to both sides to get one equals two x squared.
Dividing both sides by two gives us one half equals x squared and then if we take the square root of both sides to solve for x,
we’ll get one over the square root of two equal to x.
And actually this is positive or negative. We take the square root of the whole thing.
But remember that the square root of one is just one.
So we’ll only need one in the numerator, the square root of two we leave as the square root of two and of course, positive or negative.
So positive or negative one divided by the square root of two.
So those points on our graph, just to give you a visual, actually occur at these points here, which makes sense kind of visually.
This is where the tangent line of the graph would be completely horizontal.
So this would be the tangent line here to the graph at these points.
So that’s where the tangent lines exist and that means that the point right here, the point here and the point here,
this x coordinate would be negative one over the square root of two, and at this point,
we’ll have positive one over the square root of two.
Now we could plug in those coordinates to our original function f of x to find this exact point right here.
If we did that... we won’t do it now ‘because it’s really not helpful to the problem.
But if we did that we would get negative one over the square root of two and negative one half
and this over here would be positive one over the square root of two and positive one half.
So those would be the points where the tangent line intersects the graph and is horizontal.
So that’s how we find the points where we have horizontal tangent lines.
Vertical tangent lines are going to exist where the derivative is undefined.
So in that case, remember, here’s our derivative, this fraction here.
This fraction is only going to be undefined when the denominator is equal to zero.
Remember that we can’t divide by zero. The denominator of a fraction can never be zero.
So the slope is undefined when the tangent line is vertical, which means that if we set the denominator equal to zero,
the square root of one minus x squared, if we set that equal to zero and solve for x, we’ll find the point at which we have a vertical tangent line.
So let’s go ahead and square both sides to get one minus x squared over here, we’ll still have zero on the right.
If we add x squared to both sides, we’ll get one equals x squared.
Taking the square root of both sides, we’ll get positive or negative one is equal to x.
So now, we can see that our vertical tangent lines are going to exist at positive and negative one right here.
So right at the end of the range and the vertical tangent lines are going to come and intersect the graph right at these end points here.
So again, we could plug in positive and negative one to our original function to see what value we would get but if we did that,
we would see that the tangent line intersects the graph at the point (1, 0) over here and at the point (-1, 0) over here.
So in other words, right along the x axis.
But those are the points where the vertical tangent lines intersect the graph
and now that we know that the function has these two vertical tangent lines,
we can make a statement about the differentiability of the function.
Remember that the function is not differentiable if it’s not continuous,
it’s got a sharp point in the graph, it’s like created by an absolute value, something like that,
and also, it’s not differentiable where the tangent line is vertical.
So since we know we have two vertical tangent lines, we know that the function is not differentiable with
x equal to positive or negative one because at those two points, we have vertical tangent lines.
But everywhere else on this range, the function is differentiable. It’s continuous. It’s differentiable.
It’s just not right at the end points at this range positive and negative one.
So we can say for our final answer, we can give the location of the horizontal and vertical tangent lines.
I’m trying to do two things at once -- differentiable everywhere except x equal to positive or negative one.
So that would be the statement we could make about differentiability.
That's it. I hope that video helped you guys and I will see you in the next one.
Bye!