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Calculus 1
Course: Calculus 1 > Unit 3
Lesson 9: Second derivativesSecond derivatives (implicit equations): find expression
AP.CALC:
FUN‑3 (EU)
, FUN‑3.F (LO)
, FUN‑3.F.1 (EK)
, FUN‑3.F.2 (EK)
Given an implicit equation in x and y, finding the expression for the second derivative of y with respect to x.
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Isn't it possible to rewrite the second derivative as 4 / y^3 because if you create a common denominator you get (y^2 - x^2) / y^3? And since we know from the function that y^2 - x^2 = 4, we can substitute 4.(30 votes)
Yes, that works and is clearly a better form for the solution — good job!(15 votes)
- I tried to use the quotient rule but I got a different result. Is it me doing it incorrectly or that it has to be done as Sal did?
What I did was the same except the fact that I used quotient rule, which is as follows:
1.y - x.1 / y^2(6 votes)
Remember that we're differentiating with respect to 𝑥, which means that the derivative of 𝑦 is 𝑑𝑦∕𝑑𝑥, not 1.
So, applying the quotient rule, we get
𝑑²𝑦∕𝑑𝑥² = (1・𝑦 − 𝑥・𝑑𝑦∕𝑑𝑥)∕𝑦² = 1∕𝑦 − (𝑥∕𝑦²)・𝑑𝑦∕𝑑𝑥
and since 𝑑𝑦∕𝑑𝑥 = 𝑥∕𝑦, we get
𝑑²𝑦∕𝑑𝑥² = 1∕𝑦 − (𝑥∕𝑦²)・(𝑥∕𝑦) = 1∕𝑦 − 𝑥²∕𝑦³(16 votes)
At1:05and3:42why do we multiply by the derivative?
I see that this probably has something to do with it being a "y". But I'm just not really sure why we use it.(7 votes)
It's because of the chain rule, and because y is a function of x. If we have a function like f(g(x)), the derivative is f'(g(x))·g'(x). This is the same thing, with the outer function f(x)=x² (at1:05anyway) and the inner function g(x)=y(x).(9 votes)
What if x and y were raised to a fractional power instead of an integer? For example, what if you had x^1/3 + y^1/3 = 4? I know how to take the derivative of x^1/3 (power rule), but I'm having trouble with how I should take the derivative of y^1/3.(4 votes)
Just differentiate the exact same way as x^1/3 but multiply the dy/dx at the end. So it would just be 1/3*y^(-2/3)*dy/dx(6 votes)
If I solve the equation for y before taking the first derivative, I get the result
±( (x) / (√(x² + 4)) )
Does this result have any meaning?(4 votes)- Sure, and if you take the second derivative you will get a complicated equation that is equivalent to what is shown in the video.(5 votes)
- So when do we know to use the implicit equation and not the traditional methods?(3 votes)
- You kind of use the same method no matter what. When you have an equation you take the derivative of both sides then use algebra to find what dy/dx is.
USUALLY y is by itself on one side, and the derivative of y is dy/dx, so no algebra is necessary in that case.
Then once you have dy/dx it's pretty simple to find the second and above derivative. Does that help?(4 votes)
i kinda don't understand why i have to treat Y as a VARIABLE whilst taking the derivative of the whole equation with respect to X.(3 votes)
In implicit function, both x and y are used as variables. However, they are not used in the same way x and y are used in explicit functions, where y is entirely dependent upon x. Implicit functions simply map all the points (x,y) in which the function is true. So the function is dependent upon x and y, thus we must treat both like variables.(3 votes)
1:29Could you have instead applied the derivative operator to both sides of the equation before solving for dy/dx?
d/dx[ 2y(dy/dx) - 2x ] = d/dx[0](4 votes)
At0:53sal ways "that we are going to use the chain rule here" and so he takes the derivative of (y^2) with respect to "y" and gets 2y, and then multiplies it by the the derivative of (y^2) with respect to "x" which is just (dy/dx), and then he says that the derivative of (x^2) with respect to "x" is 2x.
So what I don't get, is how he used the chain rule there, and why doesn't he do the same thing for (x^2)?
It almost seems like he's changing the value.
I've seen lot's of examples like this, and I thought I understood, but now I'm not so sure.....
Please answer, thanks in advance,
- ncochran2(2 votes)- When we do implicit differentiation, we say that one of the variables is a function of the other. In this case, we are saying that y is a function of x. We are looking for dy/dx, which is the derivative with respect to x. To do this, we take the derivative with respect to x of both sides (that's what the d/dx means). We already know the derivative with respect to x of x^2 - It's 2x! There's not need to use the chain rule here because we are taking the derivative with respect to x. However, when we take the derivative of y^2 with respect to, we must use the chain rule. Why? Because y is a function of x! So, we take the derivative of y^2with respect to y first, and then we can multiply it by the derivative of y with respect to x.
Another thing to note: if we did want to use the chain rule for x^2, you technically could. You take the derivative of x^2 with respect to x, which is 2x, and multiply it by the derivative of x with respect to x. However, notice that the derivative of x with respect to x is just 1! (dx/dx = 1). So, this shouldn't change your answer even if you choose to think about the chain rule.(4 votes)
How would I graph these derivatives and second derivatives? Since the equations have both x and y in them, would I have to isolate y before graphing? Or would I simply replace dy/dx with y? Thanks for the help!(2 votes)- You would isolate dy/dx on the left, and have some expression in x and y on the right. Then you would replace each y with the original expression in x, leaving you with
dy/dx=(expression in x).
Then plot dy/dx on the vertical axis and x on the horizontal axis.(4 votes)
1:05Video transcript
- [Instructor] Let's say
that we're given the equation that y squared minus x squared is equal to four. And our goal is to find
the second derivative of y with respect to x, and we want to find an expression for it
in terms of x's and y's. So pause this video, and see
if you can work through this. All right, now let's do it together. Now, some of you might
have wanted to solve for y and then use some traditional techniques. But here, we have a y squared, and so it might involve a
plus or a minus square root. And so some of y'all
might have realized, hey, we can do a little bit of
implicit differentiation, which is really just an
application of the chain rule. So let's do that. Let's first find the first derivative of y with respect to x. And to do that, I'll
just take the derivative with respect to x of both
sides of this equation. And then what do we get? Well, the derivative with
respect to x of y squared, we're gonna use the chain rule here. First, we can take the derivative of y squared with respect to y, which is going to be equal to two y, and then that times the
derivative of y with respect to x. Once again, this comes
straight out of the chain rule. And then, from that, we will subtract, what's the derivative of x
squared with respect to x? Well, that's just going to be two x. And then last, but not least, what is the derivative of a
constant with respect to x? Well, it doesn't change, so it's just going to be equal to zero. All right, now we can solve for our first derivative
of y with respect to x. Let's do that. We can add two x to both sides, and we would get two y
times the derivative of y with respect to x is equal to two x. And now I can divide both
sides by two y, and I am going to get that the derivative of y with respect to x is equal to x, x over y. Now, the next step is
let's take the derivative of both sides of this with respect to x, and then we can hopefully
find our second derivative of y with respect to x. And to help us there,
actually let me rewrite this. And I always forget the quotient rule, although it might be a useful
thing for you to remember. But I could rewrite this as a product, which will help me at least. So I'm going to rewrite
this as the derivative of y with respect to x is equal to x times y to the negative one power, y to the negative one power. And now, if we want to
find the second derivative, we apply the derivative operator on both sides of this equation, derivative with respect to x. And our left-hand side is exactly what we eventually wanted to get, so the second derivative
of y with respect to x. And what do we get here
on the right-hand side? Well, we can apply the product rule. So first, we can say the
derivative of x with respect to x, well, that is just going to
be one times the other thing, so times y to the negative one power, y to the negative one power. And then we have plus x times the derivative
of y to the negative one. So plus x, what's the, times, what's the derivative of y
to the negative one power? Well, first, we can find the derivative of y to the negative one
power with respect to y. We'll just leverage the power rule there. So that's going to be negative one times y to the negative two power. And then we would multiply that times the derivative
of y with respect to x, just an application of the chain rule, times dy/dx. And remember, we know what the derivative of y with respect to x is. We already solved for that. It is x over y. So this over here is going to be x over y. And so now we just have to
simplify this expression. This is going to be equal to, and I'll try to do it part by part, that part right over there is
just going to be a one over y. And then all of this business, let's see if I can simplify that. This negative is going to
go out front, so minus, and then I'm going to have
x times x in the numerator. And then it's going to
be divided by y squared and then divided by another y. So it's going to be minus x squared over y to the third, over y to the third, or
another way to think about it, x squared times y to the negative three. And we are done. We have just figured out
the second derivative of y with respect to x in terms of x's and y's.