Why does the partial of 2y^2e^(xy^2) equal 4ye^(xy^2)?

[prace]

$$\frac{\partial_P}{\partial_y}(2ysinxcosx-y+2y^2e^{(xy^2)}$$

I worked the first part no problem, but the second part I needed a little help from my calculator. This is what I got:

$$2sinxcosx-1+4ye^{(xy^2)}$$

My question is, why does the partial of $$2y^2e^{(xy^2)}$$ come out to $$4ye^{(xy^2)}$$?

I see the where the 4y comes from, but how come the $$e^{(xy^2)}$$ stays exactly the same.

I also know that $$\frac{d}{dx}(e^x) = e^x$$

Thanks!

 

[dextercioby]

It's a mistake in the last term of the sum.

$$\frac{\partial}{\partial y} 2y^2 e^{xy^2}$$

should be done using product rule and chain rule.

Daniel.