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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

Andrew
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Danilo
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  • To answer the value of $f'$, first you have to answer what is $f'$ in a multivariable sense? – Daryl Sep 29 '12 at 22:43

3 Answers3

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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.

Zach L.
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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 $G(x,y)$.

cheepychappy
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  • I don't think I am convinced by your conclusion about the values of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial x}$. $(2,1)$ is the point where the function and gradient are being evaluated, thus $x=2$, $y=1$ – Valentin Sep 30 '12 at 00:15
  • Thanks, it was a careless mistake. Edited the original to reflect my original intention. – cheepychappy Sep 30 '12 at 04:19
  • 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 for the form of the gradient as well. Just wanted to double-check my work. – cheepychappy Sep 30 '12 at 05:38
  • For clarity: By definition $\nabla f(u,v) = \langle f_u(u,v), f_v(u,v)\rangle$ , so therefore we have been given that $f_u(2,1)=4$ and $f_v(2,1)=3$.
    Now let $w=xy^2, u=x^2+y^2, v=xy$, so $G(x,y)= w~f(u,v)$.

    Then apply the product and chain rule: $$\begin{align}G_x(x,y) &= {[w~f(u,v)]}'_x\&= w_x~f(u,v)+w~{[f(u,v)]}'_x\&= w_x~f(u,v)+w~\big(u_x~f_u(u,v)+v_x~f_v(u,v)\big)\end{align}$$

     – Graham Kemp May 25 '22 at 05:32
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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 $$\nabla G\left(1,1\right)=3\left(2\boldsymbol{i}+1\boldsymbol{j}\right)+4\boldsymbol{i}+3\boldsymbol{j}=10\boldsymbol{i}+6\boldsymbol{j}$$

Valentin
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