# Thread: Is this right? differential

1. ## Is this right? differential

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:

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

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

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.

2. ## Re: Is this right? differential

Originally Posted by Ulysses
Is this right?
Yes, it is correct (only one mistake: $\frac{\partial u}{\partial x}(0,-1)$ it is $1$ ).