Is this Differential Calculus Exercise Correct?

[Telemachus]

Hi there, I have this exercise, I'd like to know what you think, if I did this right:

If $$z=F(u,v)=5u^2+4v-7$$, with $$\begin{Bmatrix} x+y^3+u^3+v=0\\x^3+y-4u+v^4=0\end{matrix}$$. Determine if its possible the dF over a point adequately chosen.

So I choose $$P_0(0,-1,0,1)$$ which satisfies the system

Then I've verified that $$F_1,F_2$$ satisfies the implicit function theorem, both functions are polynomials of class $$C^k$$ y

$$\frac{{\partial (F_1,F_2)}}{{\partial (u,v)}}=4\neq{0}$$ then $$\exists{ E_r(P_0)}:\begin{Bmatrix} u=u(x,y)\\v=v(x,y)\end{matrix}$$

Then I consider:

$$dF=\frac{{\partial F}}{{\partial x}}dx+\frac{{\partial F}}{{\partial y}}dy$$

$$\frac{{\partial F}}{{\partial x}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial x}}$$

$$\frac{{\partial F}}{{\partial y}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial y}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}$$

And I set:

$$\frac{{\partial F_i}}{{\partial x}}= 1+3u^2\frac{{\partial u}}{{\partial x}}+\frac{{\partial v}}{{\partial x}}=0\\3x^2-4\frac{{\partial u}}{{\partial x}}+4v^3\frac{{\partial v}}{{\partial x}}=0\end{matrix}$$

From where I get:
$$\frac{{\partial u}}{{\partial x}}=0$$
$$\frac{{\partial v}}{{\partial x}}_{P_0}=-1$$

Analogous procedure for the derivatives with respect to y, and:

$$dF=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}dx+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}dy$$

Is this right?
Bye.

 

[lippyka]

Yes, that looks correct. Keep in mind that when you are computing the partial derivatives, you need to use the values of u and v at the point P_0. It should also be noted that you can simplify your work by noting that if u=0 at P_0, then its partial derivatives with respect to x and y will be 0.