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
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?
2 answers
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
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
0 answers