I Have Trouble with Using Chain Rule to Find Gradient Based on Two Given Multivariable Functions

Question

I'm taking multivariable calculus, and I'm stuck on this problem that wants me to find $\nabla f(x,y)$ at the points (x,y)=(1,0) and (x,y)=(-1,0). The two equations I'm given are $f(x,0)=x^3+3x$ for all x (eq. 1), and $f(x,1-x^2)=x^4-x^2+4x$ for all x (eq. 2). The solution uses the multivariable chain rule (we haven't covered matrices yet) to get $\nabla f(1,0)=(6,0)$ and $\nabla f(-1,0)=(6,-2)$. I'm already familier with the chain rule, parametric equations, and partial derivatives, but I'm confused with the reasoning behind using $f_x(1,0)$ and $f_x(-1,0)$ derived from eq. 1 rather than use these partial derivatives from eq. 2. To my knowledge, if I plug in $x=1$ and $x=-1$ into eq. 2, I get $f(1,0)$ and $f(-1,0)$ respectively, which from there I can compute $f_x(1,0)$ and $f_x(-1,0)$ to get $4(1)^3-2(1)+4=6$ and $4(-1)^3-2(-1)+4=2$ respectively. The $f_x(1,0)$ value is the same as eq. 1, but eq. 1's $f_x(-1,0)$ value is 6, which is different from eq. 2's $f_x$ value (2). I have no problem with the rest of the solution, as I can find $f_y$ by plugging in the other values into the chain rule formula to ultimately find $\nabla f(x,y)$.

Am I simply misunderstanding a concept here? I obviously know that I can't directly find $f_y$ from either equation by itself, but I thought that if I can find $f_x$ for eq. 1, I could also find it for eq. 2? This is my first ever question on this forum, and I would really appreciate any help!

p.s. I don't know matrices yet.

Answer 1

$f$ is a function defined on the $(x,y)$ plane and going into $\Bbb R$. You do not know what that function is. What you do know is there are a couple curves $$\begin{align}\alpha(t) &= (x_\alpha(t), y_\alpha(t)) = (t, 0)\\\beta(t) &= (x_\beta(t), y_\beta(t)) =(t,1-t^2)\end{align}$$ and you are given the values of $f$ along those curves: $$f\circ\alpha(t) = t^3 + 3t\tag{1}$$$$f\circ\beta(t) = t^4 - t^2 + 4t\tag 2$$ These two curves happen to agree at two points: $$\alpha(-1) = \beta(-1) = (-1,0)\\\alpha(1) = \beta(1) = (1,0)$$ and the problem is to find the gradient of $f$ at those two intersections. Fortunately, the tangent vectors to $\alpha$ and $\beta$ at these points are not parallel to each other. If they were, the problem would be unsolvable.

By the chain rule, we know that $$(f\circ\alpha)'(t) = f_x(\alpha(t))x_\alpha'(t) + f_y(\alpha(t))y_\alpha'(t) = 3t^2 + 3$$

Since $x_\alpha'(t) = 1, y_\alpha'(t) = 0$, this tells us $$f_x(\alpha(t)) = 3t^2 + 3$$

In particular $f_x(1,0) = 6, f_x(-1,0) = 6$. It tells us nothing about $f_y$.

Again by the chain rule $$(f\circ\beta)'(t) = f_x(\beta(t))x_\beta'(t) + f_y(\beta(t))y_\beta'(t) = 4t^3 -2t + 4$$ Now $x_\beta'(t) = 1, y_\beta'(t) = -2t$ So $$f_x(\beta(t)) - 2tf_y(\beta(t)) = 4t^3 -2t + 4$$

For our two points of interest, $$f_x(1,0) - 2f_y(1,0) = 6\\f_x(-1,0) + 2f_y(-1,0) = 2$$

But we already know what $f_x(1,0)$ and $f_x(-1,0)$ are. So $$6 - 2f_y(1,0) = 6\\6 + 2f_y(-1,0) = 2$$ And thus $$f_y(1,0) = 0\\f_y(-1,0) = -2$$

The key point to your questions is that $f(x,y), f_x(x,y), f_y(x,y)$ are functions of $(x,y)$, not $t$ (which is why I chose a different letter $t$ for the parameter, even though $x = t$ on both curves). These three functions are functions of the point, not the parameter of some curve passing through that point. Thus the $f_x(1,0)$ and $f_x(-1,0)$ determined from equation $(1)$ are the same $f_x(1,0)$ and $f_x(-1,0)$ in the calculations coming from equation $(2)$.