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March 27, 2017

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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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