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

1 Answer
Mar 24, 2015

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

By implicit differentiation:
#d/(dx)(x^2y^2)=2xy^2+2x^2y(dy)/(dx)#

We are assume that #y# is some function or functions of #x#. Think of it as

#x^2y^2=x^2*("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"#

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

So
#d/(dx)(x^2y^2)=2xy^2+x^2 2y(dy)/(dx)=2xy^2+2x^2y(dy)/(dx)#

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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^2#, then you will not need the product rule.

#(del)/(del x)(x^2y^2)=2xy^2#
because #y^2# is constant relative to #x#

and

#(del)/(del y)(x^2y^2)=2x^xy#
because #x^2# is constant relative to #y#