Let $f: \Bbb R^2 \to \Bbb R^3$ defined by $f(x,y) = (x^2 - 5y^2,ye^{2x},2x - \log(1+y^2))$ Suppose that $g:\Bbb R^2 \to \Bbb R^2$ is of class $C^1$

Question

Let $f: \Bbb R^2 \to \Bbb R^3$ defined by $f(x,y) = (x^2 - 5y^2,ye^{2x},2x - \log(1+y^2))$

Suppose that $g:\Bbb R^2 \to \Bbb R^2$ is of class $C^1$ and $g(1,2) =(0,0)$ and $$ Dg(1,2)=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $$ then $D(f \circ g)(1,2) = $

1. $$\begin{pmatrix} -15 & -20 \\ 4 & 6 \\ 2 & 4 \end{pmatrix} $$

2. $$\begin{pmatrix} -15 & -20 \\ 3 & 4 \\ 2 & 4 \end{pmatrix} $$

3. $$\begin{pmatrix} -15 & 4 & 2 \\ -20 & 6 & 4 \end{pmatrix} $$

4. $$\begin{pmatrix} -15 & 4 & 2 \\ -20 & -6 & 4 \end{pmatrix} $$

My Attempt:

Here range of $g$ contained in domain of $f$, so $f\circ g$ is well defined. Also $g(1,2) =(0,0)$ and $f(1,0) =(1,0,2)$ and $f(0,1) =(-5,1,-\log 2)$ where $(1,0)$ and $(0,1)$ are standard basis of $\Bbb R^2$

I don't know how to solve this question. Please help me.

Answer 1

Hopefully, you know the chain rule. Otherwise it could be really difficult. It holds $$ \begin{align*} D(f\circ g)(1,2) &= Df(g(1,2))\cdot Dg(1,2) \end{align*} $$ You need to calculate $Df=\begin{pmatrix}\frac{\partial f_1}{\partial x} &\frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial x}&\frac{\partial f_2}{\partial y}\\\frac{\partial f_3}{\partial x}&\frac{\partial f_3}{\partial y}\end{pmatrix}$ and evaluate it at the point $(0,0)$. Then you have to multiply it with $Dg(1,2)=\begin{pmatrix}1&2\\3&4\end{pmatrix}$.

Observe, that $f\circ g$ maps from $\mathbb R^2$ to $\mathbb R^3$. So you get the number of rows and columns of $D(f\circ g)$ .