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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 3

Lesson 8: Calculating higher-order derivatives

# Second derivatives (implicit equations): find expression

Given an implicit equation in x and y, finding the expression for the second derivative of y with respect to x.

## Want to join the conversation?

• 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.
• Yes, that works and is clearly a better form for the solution — good job!
• 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
• 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
• At and why 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.
• 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 (at anyway) and the inner function g(x)=y(x).
• 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.
• 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
• 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?
• Sure, and if you take the second derivative you will get a complicated equation that is equivalent to what is shown in the video.
• So when do we know to use the implicit equation and not the traditional methods?
• 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?
• 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.
• 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.
• Could 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]
• At sal 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.....
- ncochran2
• 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.