Posted by **Samantha** on Monday, November 3, 2008 at 8:59pm.

Can someone explain the method of implicit differentiation to me.

I know that we differentiation both sides of the equation with respect to x (I don't understand what they mean by with respect to x). And then y is a function of x...I don't understand what that means.

For example,

Find y'' if x^4+y^4 = 16

Basically we need to differentiate both sides..as far as I know..

so ..(for y')

4x^3+4y^3 = 0

BUT..

in the textbook it's

4x^3+4y^3y' = 0

Where does the y' come from?

- Calculus -
**Damon**, Monday, November 3, 2008 at 9:31pm
That means take d/dx of everything

x^4 + y^4 = 16

d/dx x^4 = 4 x^3 dx/dx = 4 x^3

d/dx y^4 = 4 y^3 dy/dx

d/dx 16 = 0

so

4 x^3 + 4 y^3 dy/dx = 0

dy/dx = -x^3/y^3

- Calculus -
**Damon**, Monday, November 3, 2008 at 9:36pm
basic thing is

d/dx u^n = n u^(n-1) du/dx

when u happens to be x, you get dx/dx at the end which is 1

but if u happens to be y, then you have that dy/dx at the end

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