What is the derivative of x^2y^2x2y2?

1 Answer
Mar 24, 2015

I will assume that you want d/(dx)(x^2y^2)ddx(x2y2)

By implicit differentiation:
d/(dx)(x^2y^2)=2xy^2+2x^2y(dy)/(dx)ddx(x2y2)=2xy2+2x2ydydx

We are assume that yy is some function or functions of xx. Think of it as

x^2y^2=x^2*("some function")^2x2y2=x2(some function)2

To differentiate this, you'd need the product rule and the chain rule.

d/(dx)[x^2*("some function")^2]=2x*("some function")^2+x^2*2("some function")*"the derivative of the function"ddx[x2(some function)2]=2x(some function)2+x22(some function)the derivative of the function

That looks kinda complicated, but we have names for the "some function"some function and for its derivative. We call them yy and (dy)/(dx)dydx.

So
d/(dx)(x^2y^2)=2xy^2+x^2 2y(dy)/(dx)=2xy^2+2x^2y(dy)/(dx)ddx(x2y2)=2xy2+x22ydydx=2xy2+2x2ydydx

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If my assumption was mistaken:

If you wanted to find the partial derivatives of the function f(x,y)=x^2y^2f(x,y)=x2y2, then you will not need the product rule.

(del)/(del x)(x^2y^2)=2xy^2x(x2y2)=2xy2
because y^2y2 is constant relative to xx

and

(del)/(del y)(x^2y^2)=2x^xyy(x2y2)=2xxy
because x^2x2 is constant relative to yy