# Calculus

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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 -

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 -

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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