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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

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    To answer the value of $f'$, first you have to answer what is $f'$ in a multivariable sense?2012-09-29

3 Answers 3

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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.

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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 $G(x,y)$.

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    Sorry, I had poorly grouped the partials into directional sums in my original definition of $\nabla f(u,v)$, fixed that too. The form I've written checks with [WolframAlpha's answer](http://www.wolframalpha.com/input/?i=del%28x^2*y*f%28x^2%2By^2%2Cxy%29%29) for the form of the gradient as well. Just wanted to double-check my work.2012-09-30
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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 $\nabla G\left(1,1\right)=3\left(2\boldsymbol{i}+1\boldsymbol{j}\right)+4\boldsymbol{i}+3\boldsymbol{j}=10\boldsymbol{i}+6\boldsymbol{j}$