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I am having great trouble trying to understand this chain rule question. As you can see, there are three equalities. $f(x,y) = f(w,w) = f(uv, u^2 + v^2)$ This makes absolutely no sense to me! When trying to calculate the partial derivative of f with respect to $x$, what use do we have of "$w$"?

Thanks a bunch if you can take to time to explain a little!

Question:

Let $z = f(x,y)$ be a differentiable function such that $$\begin{array}{ccc} f(3,3) = 1, & f_x(3,3) = -2, & f_y(3,3) = 11, \\ f(2,5) = 1, & f_x(2,5) = 7, & f_y(2,5) = -3. \end{array}$$ Suppose $w$ is a differentiable function of $u$ and $v$ satisfying the equation $$f(w,w) = f(u,v, u^2+v^2)$$ for all $(u,v)$. Find $\displaystyle \frac{\partial w}{\partial u}$ at $(u,v,w) = (1,2,3)$.

Proposed Solution:

Differentiating the identity $$f(w,w) = f(uv,u^2+v^2)$$ with respect to $u$ gives $$f_x(w,w)\frac{\partial w}{\partial u} + f_y(w,w) \frac{\partial w}{\partial u} = f_x(uv, u^2+v^2) \frac{\partial (uv)}{\partial u} + f_y(uv, u^2+v^2) \frac{\partial (u^2+v^2)}{\partial u}$$ by the Chain Rule. Hence $$\left(f_x(w,w) + f_y(w,w)\right) \frac{\partial w}{\partial u} = f_x(uv, u^2+v^2)v + f_y(uv, u^2+v^2) 2u$$ which leads to $$\left(f_x(3,3) + f_y(3,3)\right) \frac{\partial w}{\partial u} = 2f_x(2,5) + 2f_y(2,5)$$ after substituting $(u,v,w) = (1,2,3)$. Now using $f_x(3,3) = -2$, $f_y(3,3) = 11$, $f_x(2,5) = 7$, and $f_y(2,5) = -3$, we conclude that $$\frac{\partial w}{\partial u} = \frac{8}{9} \quad \text{at} \quad (u,v,w) = (1,2,3).$$

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  • $\begingroup$ We compute the partial derivative of $f $ at $(x ,y)= (w,w)$ where $w$ is a differentiable function of $u$ and $v$ $\endgroup$
    – 19aksh
    Commented May 19, 2019 at 17:18

2 Answers 2

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Here is a simpler example:

$$f(x,y) = (x+y)^2.$$

If we plug in $w$ for both $x$ and $y$ we get

$$f(w,w) = (2w)^2 = f(x,y) \iff w = \frac{x+y}{2}$$

(assuming $x, y \geq 0$).

So, for the right choice of $w = w(x,y)$ (i.e. as a function of $x$ and $y$), we do indeed have $f(x,y) = f(w,w)$. Your example is made complicated by the fact that you also have $u$s and $v$s, so you need to select the right $u$, $v$ and $w$ to make all the equalities match. (But you don't need to worry about what $w$ "looks like" precisely for the question, just assume that there is one.)

As for the chain rule, you have to differentiate the composition. If you have $f(g(t,s), h(t,s))$, then, differentiating with respect to $t$ gives:

$$ \frac{\partial f}{\partial t}(\underbrace{g(t,s)}_{x=g(t,s)},\underbrace{h(t,s)}_{y=h(t,s)}) = \frac{\partial f}{\partial x}\frac{\partial g}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial h}{\partial t} $$.

The $x$ and $y$ in the $f$ derivatives indicate that we differentiate with respect to the argument in the $x$-position of $f(x,y)$ (so $g$) and the $y$-position of $f(x,y)$ (which is $h$).

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  • $\begingroup$ What I am confused with is that in the equality 𝑓(𝑤,𝑤)=𝑓(𝑢𝑣,𝑢^2+𝑣^2), we only have u, v, w. Why do we have to take the derivative with respect to x? $\endgroup$
    – Skkk
    Commented May 19, 2019 at 18:19
  • $\begingroup$ It's not "with respect to $x$", it's "with respect to the first argument to the function". $\endgroup$
    – ryan221b
    Commented May 19, 2019 at 18:20
  • $\begingroup$ So by "first argument to the function", you mean the first term inside the brackets, right? That makes some sense now. $\endgroup$
    – Skkk
    Commented May 19, 2019 at 18:22
  • $\begingroup$ I hope the edit makes it clearer :) $\endgroup$
    – ryan221b
    Commented May 19, 2019 at 18:26
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It's easy to get lost in a thicket of variables. Perhaps it would help to go back to the definition of derivative and derive the general formula.

Let $g:\mathbb R^2\to \mathbb R^2$ be defined by $(x,y)\mapsto (xy, x^2 + y^2)$. Then, $F:\mathbb R^2\to \mathbb R$ is a composition $F(x,y)=(f\circ g)(x,y).$

Now, apply the chain rule for derivatives to $F$:

$F'(x,y)=f'(g(x,y))\circ g'(x,y).$

Now, $f'(g(x,y))$ and $g'(x,y)$ are linear transformations, which can be expressed as $1\times 2$ and $2\times 2$ matrices, respectively:

$f'(g(x,y))=\begin{pmatrix} \frac{\partial f}{\partial x}(g(x,y))) &\frac{\partial f}{\partial y}(g(x,y)) \end{pmatrix}=\begin{pmatrix} \frac{\partial f}{\partial x}((xy, x^2 + y^2)) &\frac{\partial f}{\partial y}((xy, x^2 + y^2)) \end{pmatrix}$

$g'(x,y)=\begin{pmatrix} \frac{\partial g_1}{\partial x}(x,y) & \frac{\partial g_2}{\partial x}(x,y)\\ \frac{\partial g_1}{\partial y}(x,y) & \frac{\partial g_2}{\partial y}(x,y) \end{pmatrix}=\begin{pmatrix} y & 2x\\ x & 2y \end{pmatrix}$.

Therefore,

$\begin{pmatrix} \frac{\partial F}{\partial x}(x,y)&\ \frac{\partial F}{\partial y}(x,y) \end{pmatrix}=\begin{pmatrix} \frac{\partial f}{\partial x}((xy, x^2 + y^2)) &\frac{\partial f}{\partial y}((xy, x^2 + y^2)) \end{pmatrix}\cdot \begin{pmatrix} y & 2x\\ x & 2y \end{pmatrix}=\begin{pmatrix} y\frac{\partial f}{\partial x}((xy, x^2 + y^2)) +x\frac{\partial f}{\partial y}((xy, x^2 + y^2))&\ 2x\frac{\partial f}{\partial x}((xy, x^2 + y^2)) +2y\frac{\partial f}{\partial y}((xy, x^2 + y^2)) \end{pmatrix}$.

All that remains now is to substitute the values into these last expressions.

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