Derivative of $\ z=v^{3}u^{5} $ by chain rule and substitution Let $\ z=u^{3}v^{5} $ where $\ u=x+y, v=x-y $ Find $\ \frac{dz}{dy} $
For that I just did
$$\ \frac{dz}{dy}=\frac{dz}{du}\frac{du}{dy}+\frac{dz}{dv}\frac{dv}{dy} $$
And I got: $$\ 3(x+y)^{2}(y-5)^{5}+5(x+y)^{3}(x-y)^{4}$$ Is this right?
This is the solution by the chain rule, right?
But then the problem says I need to make substitution and explicit computation, but I have no idea of this, It was not covered in the lectures. 

$$\frac{dz}{du}=3u^2v^5$$ and $$\frac{du}{dy}=1$$ because for $\frac{du}{dy}$ you treat $x$ as a constant. Now you substitute the expressions for $u$ and $v$.
The second part is calculated similarly.
