How do i find $\partial z/ \partial u$ when $u=0$, $v=1$ if $z=\sin xy+x\sin y$, $x=u^2+v^2$,$y=uv$.

Question

The chain rule states that the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x).$ ... For example, $\sin(x^2)$ is a composite function because it can be constructed as $f(g(x))$ for $f(x)=\sin(x)$ and $g(x)=x^2$.

Answer 1

Since $x, y$ are functions of $u$, $\frac{dz}{du} = y\cos(xy)\frac{dx}{du} + x\cos(xy)\frac{dy}{du} + \frac{dx}{du} \sin y + x\cos y \frac{dy}{du}$ You will know that $\frac{dx}{du} = 2u, \frac{dy}{du}=v$

Answer 2

Hint :$$ \frac{\partial{z}}{\partial{u}}=\frac{\partial{z}}{\partial{x}}\frac{\partial{x}}{\partial{u}}+\frac{\partial{z}}{\partial{y}}\frac{\partial{y}}{\partial{u}} $$