Find the differential (if exist) of the function $h(\vec{x}) = \frac{f^3(\vec{x})+f(\vec{x})g^2(\vec{x})}{f^2(\vec{x})+g(\vec{x})}.$

Question

Let $A\subset\mathbb{R}^n$ a noempty open set and $\vec{x}_0\in A$. Let $f,g:A\to\mathbb{R}$ two differentiable function in $\vec{x}_0$ so that, $g(\vec{x})>0$, $\forall \vec{x}\in A$. Consider the function $$h:A\to\mathbb{R}\ ,\ \vec{x}\mapsto h(\vec{x}) = \frac{f^3(\vec{x})+f(\vec{x})g^2(\vec{x})}{f^2(\vec{x})+g(\vec{x})}.$$ Show that $h$ is differentiable in $\vec{x}_0$ and calcule $Df(\vec{x}_0)$.

Answer 1

We can write that $h\left(\vec{x}\right)=\psi\circ\phi\left(\vec{x}\right)$ with $$\begin{array}{ccccc} \phi & : & \mathbb{R}^n & \longrightarrow & \mathbb{R}^2\\ & & \vec{x} & \longmapsto & \left(f(\vec{x})^3+f(\vec{x})g(\vec{x})^2,f(\vec{x})^2+g(\vec{x})\right) \end{array}$$ and $$\begin{array}{ccccc} \psi & : & \mathbb{R}^2\setminus\left(\mathbb{R}\times\{0\}\right) & \longrightarrow & \mathbb{R}\\ & & \left(u,v\right) & \longmapsto & \frac{u}{v} \end{array}.$$ This is consistent because $f(\vec{x})^2+g(\vec{x})>0$. Now we have $$\mathrm{d}\phi\left(\vec{x}_0\right)[\vec{y}]=\left(3f\left(\vec{x}_0\right)^2\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]+\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]g\left(\vec{x}_0\right)^2+2f\left(\vec{x}_0\right)g\left(\vec{x}_0\right)\mathrm{d}g\left(\vec{x}_0\right)[\vec{y}],2f\left(\vec{x}_0\right)\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]+\mathrm{d}g\left(\vec{x}_0\right)[\vec{y}]\right)\quad\quad\quad\forall\vec{y}\in\mathbb{R}^n\quad\quad\quad(1)$$ and $$\mathrm{d}\psi\left(u_0,v_0\right)[h,k]=\frac{h}{v_0}-\frac{u_0k}{v_0^2}\quad\quad\quad\forall\left(h,k\right)\in\mathbb{R}^2\quad\quad\quad(2).$$ Hence, $h$ is differentiable at the point $\vec{x}_0$ and its differential is given by the well known formula $$\mathrm{d}h\left(\vec{x}_0\right)[\vec{y}]=\mathrm{d}\psi\left(\phi\left(\vec{x}_0\right)\right)\circ\mathrm{d}\phi\left(\vec{x}_0\right)[\vec{y}]\quad\quad\quad\forall\vec{y}\in\mathbb{R}^n$$ that is, $\left(u_0,v_0\right)$ in $(2)$ are replaced by $\left(f(\vec{x}_0)^3+f(\vec{x}_0)g(\vec{x}_0)^2,f(\vec{x}_0)^2+g(\vec{x}_0)\right)$ and $\left(h,k\right)$ in the same formula $(2)$ becomes the (long) vector of $\mathbb{R}^2$ in $(1)$.