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.
Best Answer
Demo.
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]. $\blacktriangleright$