a) Let A=\{(u,v)\in\mathbb R^2:u>0,v>0\} and g:A\to\mathbb R^2, defined by g(u,v)=\left(\ln(uv),\frac1{u^2+v}\right). Prove that around (1,1), g has differentiable inverse function h such that (1,1)=h\left(0,\frac12\right). Compute h'\left(0,\frac12\right).

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


We have

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

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