If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Differential Calculus

### Course: Differential Calculus>Unit 3

Lesson 9: Second 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. • 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. • 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? • 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. • 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 • 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  