Chain Rule of Partial Derivatives If $f$ is a differentiable function defined $f: \mathbb{R}^2 \to \mathbb{R}$ and $f(2,1) = 3$ and $\nabla f(2,1) = (4,3)$, find $\nabla G(1,1)$ with $G(x,y) := x^2yf(x^2 +y^2, xy)$.
I wrote the $G_x$ as $2xyf(x^2 +y^2, xy) + x^2yf'(x^2 + y^2, xy)f_x(x^2 + y^2, xy)$ but I don't know what the value of $f'(x^2 + y^2, xy)$ is.
Thanks in advance
 A: It looks like you computed $\partial_x f(x^2 + y^2, xy)$ incorrectly. Define $h:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $h(x,y) = (x^2 + y^2, xy)$ and use the chain rule on $f(h(x,y))$ to get the correct formula, namely $2xf_x(x^2 + y^2,xy) + yf_y(x^2 + y^2, xy)$. Remember, it's the sum of the partials in each slot times the derivative of function you see in that slot.
A: If you are wondering about the inner mechanisms of the chain rule for partial derivatives, you can read about it on Paul's Online Notes, specifically under 'Case 2'. If you think of $\nabla f(x,y)$ as $\frac{\partial f}{\partial x}\boldsymbol{i} + \frac{\partial f}{\partial y}\boldsymbol{j}$, you know that at point $(2,1)$ $\frac{\partial f}{\partial x} = 4$ and $\frac{\partial f}{\partial y} = 3$. 
Next you can set $u(x,y) = x^2 + y^2$ and $v(x,y) = xy$ and use the partial derivatives $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, and $\frac{\partial v}{\partial y}$ to create a new expression...
$\nabla f(u,v) = (\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x})\boldsymbol{i} + (\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y})\boldsymbol{j}$
From the logic of the first paragraph, you know what the values of $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$ are at the point $(2,1)$. The expressions in the parentheses transform that $f(u,v)$ result in terms of the $x$ and $y$ used in $G(x,y)$.
A: Denote
$$g(x,y)=x^{2}y$$
Then by product rule
$$\nabla G=f\nabla g+g\nabla f$$
$$\nabla g=2xy\boldsymbol{i}+x^{2}\boldsymbol{j}$$
Now write
$$\begin{cases}
x^{2}+y^{2} & =2\\
xy & =1
\end{cases}$$
Multiply the second equation by 2, add and subtract from the first one obtaining respectively
$$\left(x+y\right)^{2}=4$$
$$\left(x-y\right)^{2}=0$$
Hence
$$x=y=\pm1$$
Finally
$$\nabla G\left(1,1\right)=3\left(2\boldsymbol{i}+1\boldsymbol{j}\right)+4\boldsymbol{i}+3\boldsymbol{j}=10\boldsymbol{i}+6\boldsymbol{j}$$
