# Computing the derivative of a system of equations in the neighborhood of a point using implicit differentiation and the implicit function theorem

I'm solving the following problem on an old exam in real analysis. Thus, only such methods may be used.

The system \begin{align*} \begin{cases} \sin(x+y)+\sin(y+z)+z=0 \\ \cos(x+y)+\cos(y+z)+y-2=0 \end{cases} \end{align*} is satisfied at the point $(0,0,0)$. Show that $(x,y)$ can be solved in a neighborhood of $(0,0)$ as a function of $z$ for $z$ near 0. Calling the function $f(z)$, calculate explicitly $f'(0)$.

I solved a similar problem in a similar way, but am a bit confused as how to deal with the differences here. The problem was the following.

The system \begin{align*} \begin{cases} x+y+z=6 \\ x^2+y^2+z^2=14 \end{cases} \end{align*} is satisfied at the point $(1,2,3)$. Show that $x$ and $y$ can be solved in a neighborhood of $(1,2,3)$ as a function of $z$. Calculate also $x'(3)$ and $y'(3)$, where $x$ and $y$ are regarded as functions of $z$.

My solution was the following.

Denote $F(x,y,z)=x^2+y^2+z^2-14$ and $G(x,y,z)=x+y+z-6$. The point $(1,2,3)$ is a solution to this system of equations. Let $H=(F,G)$. If \begin{align*} \frac{\partial(F,G)}{\partial(x,y)}= \begin{vmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{vmatrix} = \begin{vmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial y} \end{vmatrix} = \begin{vmatrix} 2x & 2y \\ 1 & 1 \end{vmatrix} =2x-2y\neq0 \end{align*} for $(x,y,z)=(1,2,3)$. The Implicit Function Theorem now implies that there are $C^1$ functions $x(z)$ and $y(z)$ defined on the ball $B_r(1,2,3)$ that satisfy the above system with $x(z)=2$ and $y(z)=3$.

Now \begin{align*} x+y+z&=6 \\ x^2+y^2+z^2&=14 \end{align*} becomes \begin{align*} x'+y'+1&=0 \\ 2xx'+2yy'+2z&=0 \end{align*} becomes \begin{align*} x'+y'+1&=0 \\ 2x'+4y'+6&=0 \end{align*} since $(x,y,z)=(1,2,3)$. Now \begin{align*} x'&=-y'-1 \\ y'&=-x'-1 \end{align*} are used to obtain \begin{align*} 2x'+4y'+6&=2x'+4(-x'-1)+6=2x'-4x'-4+6=-2x'+2=0\Leftrightarrow x'=1 \\ 2x'+4y'+6&=2(-y'-1)+4y'+6=-2y'-2+4y'+6=2y'+4=0\Leftrightarrow y'=-2. \end{align*}

I know how to solve for a functions $x(z)$ and $y(z)$, but not for $f(z)$. This makes it hard for me to compute the derivative there implicitly as well.

• I'm stuck with finding $f(z)$. That's kind of the problem. I'm even stuck with showing that the determinant is nonzero. Commented Aug 27, 2018 at 13:48
• Personally, these are basically the same. You do not need to explicitly write down the expression of $f \colon z \mapsto (x(z), y(z))$.
– xbh
Commented Aug 27, 2018 at 13:53
• Okay. I don't see it. Do could you write it down as an answer and I'll have a look at it? Commented Aug 27, 2018 at 13:54

Solution. $\blacktriangleleft$ Denote the LHS of two equations respectively by $F(x,y,z), G(x,y,z)$. Then \begin{align*} & \partial_x F = \cos(x+y), & &\partial_y F = \cos(x+y) + \cos(y+z),& \\ &\partial_x G = -\sin(x+y),& &\partial_y G = -\sin(x+y) - \sin(y+z) + 1,& \end{align*} where each of these partial derivatives is continuous around $(0,0,0)$. Since $$\det \left.\left( \frac {\partial(F,G)} {\partial (x,y)}\right)\right|_{(0,0,0)} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \neq 0,$$ the implicit function theorem could be applied to $F = 0, G= 0$, i.e. around $(0,0,0)$ there exists $\mathcal C^1$ functions $x(z), y(z)$ s.t. $F(x(z), y(z),z) = G(x(z), y(z), z) = 0$ and $x(0) = y(0) = 0$.
Now calculate the derivatives. Take the derivative w.r.t. $z$ in the given system of equation: $$\begin{cases} \cos(x+y) (x’ + y’) + \cos(y + z) (y’ + 1) +1 = 0 \\ -\sin(x+y)(x’+y’) - \sin(y+z) (y’ + 1) + y’ = 0 \end{cases},$$ and let $z = 0$ we have $$\begin{cases} x’(0) + 2y’(0) + 1 = 0\\ y’(0) = 0 \end{cases},$$ then $x’(0) = -1, y’(0) = 0$, hence $f’(0) = [-1\;0]^{\mathsf T}$ [maybe you do not need to take the transpose, depend on your notation system]. $\blacktriangleright$
• Shouldn't it be $\cos(x+y)(x'+y')+\cos(y+z)(y'+1)+1=0$ since the derivative of $z$ wrt $z$ is 1? Otherwise, your solution was great. Thanks! It cleared some things up for me. Commented Aug 27, 2018 at 19:27