[tex]\frac{\partial_P}{\partial_y}(2ysinxcosx-y+2y^2e^{(xy^2)}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]2sinxcosx-1+4ye^{(xy^2)}[/tex]

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

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

I also know that [tex]\frac{d}{dx}(e^x) = e^x[/tex]

Thanks!

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# Homework Help: Explanation of this Partial?

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