0

Find $f_y'(x,x^2)$ for a differentiable function $f(x,y)$ satisfying the conditions $f(x,x^2) = const, f_x'(x,x^2) = x$. I found that if $f(x,y) = \frac{1}{2}x^2 - \frac{1}{2}y$, it satisfies this conditions and $f'_y(x,x^2)$ would be $\frac{1}{2}$, but there might be another function that might satisfy the conditions and have another answer, is it? And if so, i don't know how to solve and need help.

Galep Impet
  • 45

1 Answers1

1

Define $\phi(t) = (t,t^2)$. Then, by the chain rule $$\frac{d}{dx} f(\phi(x)) = \partial_x f(x,x^2) + 2x\partial_y f(x,x^2) = x + 2x\partial_y f(x,x^2) \quad \forall x\in \mathbb{R}.$$

Since $f(\phi(x)) = const$, we have $\frac{d}{dx} f(\phi(x)) = 0$, so it must be $$\partial_y f(x,x^2) = -\frac{1}{2}$$ for all $x\neq 0$. If $f$ is continuously differentiable then also $\partial_y f(0,0) = -\frac{1}{2}$.

Marlon
  • 71