Encontre a derivada parcial fy se f(x,y) = y.senxy ?

Seja:

\(\frac{\partial }{\partial \:y}\left(y.senxy\:\right)\)

Vamos utilizar a regra do produto \(\left(f\cdot g\right)'=f'\cdot g+f\cdot g'\)

\(\frac{\partial \:}{\partial \:y}\left(y\right)\ sen \left(xy\right)+\frac{\partial \:}{\partial \:y}\left(\ sen \left(xy\right)\right)y\)

Mas:

\(\frac{\partial \:}{\partial \:y}\left(y\right)=1\\ \frac{\partial \:}{\partial \:y}\left(\ sen \left(xy\right)\right)=\cos \left(xy\right)x\)

Assim:

\(\frac{\partial \:}{\partial \:y}\left(y\right)\ sen \left(xy\right)+\frac{\partial \:}{\partial \:y}\left(\ sen \left(xy\right)\right)y=1\cdot \sin \left(xy\right)+\cos \left(xy\right)xy\)

Simplificando:

\(\ sen \left(yx\right)+yx\cos \left(yx\right)\)

Ou seja

\(\boxed{\frac{\partial }{\partial \:y}\left(y.senxy\:\right)=\ sen \left(yx\right)+yx\cos \left(yx\right)}\)