Solving a function from 2 partial derivatives

Question

I am not sure how you are supposed to calculate this:

Knowing that function $f(x,y)$ satisfies $f_x(48,-4) = 2$ and $f_y(48,-4)=-4$, and that function $g(u,v)$ is given by $g(u,v) = f(3*u^2*v^2,-u*v)$,

find $g_u(2,2)$ and $g_v(2,2)$

Answer 1

Let $$ h(x, y) = \begin{cases} \frac{xy}{x^2 + y^2} & (x, y) \ne (0,0) \\ 0 & (x,y) = (0,0) \end{cases}. $$

Then it's not too hard to prove (with limits ) that $h_x(0,0) = h_y(0,0) = 0$, but $h$ is not actually differentiable at $(0,0)$.

Now define $$ f(x, y) = h(x-48, y + 4) + 2x - 4y. $$ It's not hard to see that $f_x(48, -4) = h_x(48-48, -4 +4) + 2 = 2$, and that $f_y(48, -4) = -4$ similarly. So $f$ satisfies the hypotheses of your problem.

Now with $g(u, v) = f(3 u^2v^2, -uv)$, it's tempting to use the chain rule to say that \begin{align} g_u(2,2) &= f_x(48, -4) q_u(2,2) + f_y(48, -4) r_u(2,2)\\ &= 2 \cdot 48 + (-4) (-2)\\ &= 104 \end{align} (unless I've done the arithmetic wrong in my head). Also: I've used $q(u,v) = 3u^2v^2$ and $r(u,v) = -uv$ to denote the two functions used in defining $g$.

That's probably the answer expected for this problem. But let's work out the details.

\begin{align} g(u, v) &= f(3u^2v^2, -uv) \\ &= h(3u^2v^2 - 48, -uv + 4) + 2(3u^2v^2) - 4(-uv) & \text{by defn of $f$} \\ &= h(3u^2v^2 - 48, 4-uv ) + 6u^2v^2 + 4uv & \text{by algebra} \\ &= 6u^2v^2 + 4uv + \begin{cases} \frac{(3u^2v^2-48)(4-uv)}{(3u^2v^2-48)^2 + (4-uv)^2} & 3u^2v^2-48 \ne 0 \text{ or } 4-uv \ne 0\\ 0 & \text{otherwise} \end{cases}& \text{definition of $h$} \\ \end{align}

As it happens, at $(u,v) = (2,2)$, we have both $3u^2v^2-48 =0$ and $4-uv = 0$, so $$ g(2,2) = 6\cdot 2^2\cdot 2^2 + 4\cdot 2\cdot 2 + 0 = 112. $$ Now let's compute the partial of $g$ with respect to $u$, at the point $(2,2)$, which is defined to be \begin{align} \lim_{t \to 0} \frac{g(2+t, 2) - g(2,2)}{t}. \end{align} Right away, we can substitute in a value for $g(2,2)$, but then things get messy for a while. Note that for $t \ne 0$, we end up in the first case for the definition of $f$. With that in mind, let's just go with it: \begin{align} \lim_{t \to 0} \frac{g(2+t, 2) - g(2,2)}{t} &= \lim_{t \to 0} \frac{g(2+t, 2) - 112}{t}\\ &= \lim_{t \to 0} \frac{f(3(2+t)^2 \cdot 2^2, -(2+t)\cdot 2) - 112}{t}\\ &= \lim_{t \to 0} \frac{f(12(2+t)^2 , -2(2+t)) - 112}{t}\\ &= \lim_{t \to 0} \frac{\frac{(12(2+t)^2 ( -2(2+t))}{(12(2+t)^2)^2 + (-(2(2+t))^2} - 112}{t}\\ &= \lim_{t \to 0} \frac{-\frac{288(2+t)^3}{(12(2+t)^2)^2 + (-(2(2+t)))^2} - 112}{t}\\ &= \lim_{t \to 0} \frac{-\frac{288(2+t)^3}{144(2+t)^4 + 4(2+t)^2} - 112}{t}\\ &= \lim_{t \to 0} \frac{-\frac{288(2+t)}{144(2+t)^2 + 4} - 112}{t}\\ \end{align} As $t$ approaches $0$, the numerator approaches $$ \frac{-288 \cdot 2}{144 \cdot 4 +4} - 112 \approx -0.993 - 112 \approx -113 $$ while the denominator approaches $0$. So this limit does not exist. In short, for this particular choice of $f$, the value of $g_u(2,2)$ does not exist, rather than being the $104$ gotten by applying the chain rule willy-nilly without considering whether $f$ was in fact differentiable, rather than merely having partial derivatives.

Answer 2

Hint: $3 \cdot 2^2 \cdot 2^2 = 48$ and $-2 \cdot 2 = -4$, hence you only need to take partial derivatives of $g$ and plug in the values you know from $f$.