Thread: implicit differentiation

I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1

Originally Posted by kwivo

I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1

Implicit differentiation is really just the chain rule in disguise.

If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying


So let's take your problem:


Take the derivative:


Now solve for

-Dan

Originally Posted by kwivo

I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1

for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

if you differentiate an x-term you attach dx/dx to it (since you took the derivative of the variable x with respect to x), we usually don't write it though, because derivative notations can function as fractions and so dx/dx cancels to become 1

similarly, when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x. we often write y' when it is understood what we are differentiating with respect to

here goes:




differentiating implicitly, we get:



now solve for , which is



EDIT: Beaten by "the man." And he gave a better explanation too!

One more question: I don't understand how you get the dy/dx after the 2y. Where did that come from? Isn't 2y already the derivative of y^2.

did you read the posts? what did you not get?

Originally Posted by topsquark

If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying

Originally Posted by Jhevon

for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x.

Sorry, I overlooked that part.