# Implicit Function Theorem

• Oct 3rd 2011, 10:16 AM
Markov
Implicit Function Theorem
a) Let $\displaystyle A=\{(u,v)\in\mathbb R^2:u>0,v>0\}$ and $\displaystyle g:A\to\mathbb R^2,$ defined by $\displaystyle g(u,v)=\left(\ln(uv),\frac1{u^2+v}\right).$ Prove that around $\displaystyle (1,1),$ $\displaystyle g$ has differentiable inverse function $\displaystyle h$ such that $\displaystyle (1,1)=h\left(0,\frac12\right).$ Compute $\displaystyle h'\left(0,\frac12\right).$

b) Consider $\displaystyle h\circ f$ defined around $\displaystyle (1,-1,1).$ Prove that $\displaystyle h\circ f$ is differentiable and compute $\displaystyle (h\circ f)'(1,-1,1)$ where $\displaystyle f(x,y,z)=(\sin(st+z),(1+s^2)^{zt}).$

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We have

$\displaystyle J=\left[ \begin{matrix}\frac{1}{u} & \frac{1}{v} \\[0.2cm]-\frac{2u}{{{({{u}^{2}}+v)}^{2}}} &-\frac{1}{{{({{u}^{2}}+v)}^{2}}} \\\end{matrix} \right]\implies \left| j \right|=\frac{2u}{v{{({{u}^{2}}+v)}^{2}}}-\frac{1}{u{{({{u}^{2}}+v)}^{2}}}=\frac{2{{u}^{2}}-v}{uv{{({{u}^{2}}+v)}^{2}}},$

so for $\displaystyle (u,v)=(1,1),$ $\displaystyle g$ has inverse. I don't know how to compute $\displaystyle h'\left(0,\frac12\right).$

b) $\displaystyle h\circ f$ is differentiable because $\displaystyle h$ and $\displaystyle f$ are, now, I'm not sure $\displaystyle (h\circ f)'(1,-1,1)$ is a product of matrices, how to do it?