I struggle with this exercice.

If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is differentiable and $g: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is given by $g(x_1, x_2) = f(x^2_1- x^2_2, 2x_1x_2)$. Show that $g$ is also differentiable and compute $\frac{\partial{g}}{\partial{x_1}}$ and $\frac{\partial{g}}{\partial{x_2}}$ in terms of $\frac{\partial{f}}{\partial{x_1}}$ and $\frac{\partial{f}}{\partial{x_2}}$

Clearly, this is a compound function so we have to use the chain rule somewhere.

As we know : if $g = f \circ h$ then $Dg(a)[v] = Df(h(a))[Dh(x)[v]]$ where $a$ is an interior point of $\mathbb{R}^3$ and $v \in \mathbb{R}^3$ is a direction vector.

In class, teacher showed this :

Let $x = (x_1, x_2)$ with $x_1 = (x \cdot e_1)$ and $x_2 = (x \cdot e_2)$

$g = f \circ h$ with $h(x) = ((x \cdot e_1)^2 - (x \cdot e_2)^2)e_1 + 2(x \cdot e_1)(x \cdot e_2)e_2$

So, $\frac{\partial{g}}{\partial{x_1}} = Dg(x)[e_1] = Df(h(x))[Dh(x)[e_1]]$

From here, I don't really know what to do. I don't understand the use of a basis $(e_1, e_2)$ in differentiation and how to derive a dot product with a vector of this basis.

Thanks for the help.


Denoting by $y_1 = x_1^2-x_2^2$ and $y_2=2x_1 x_2$ we have $g(x_1, x_2)=f(y_1, y_2)$ and using the chain rule $$ \frac{\partial g}{\partial x_1} = \frac{\partial f}{\partial y_1} \frac{\partial y_1}{\partial x_1} + \frac{\partial f}{\partial y_2} \frac{\partial y_2}{\partial x_1} =2\left(x_1\frac{\partial f}{\partial y_1}+x_2\frac{\partial f}{\partial y_2}\right)\,. $$ Similarly $$ \frac{\partial g}{\partial x_2} = \frac{\partial f}{\partial y_1} \frac{\partial y_1}{\partial x_2} + \frac{\partial f}{\partial y_2} \frac{\partial y_2}{\partial x_2} =2\left(-x_2\frac{\partial f}{\partial y_1}+x_1\frac{\partial f}{\partial y_2}\right)\,. $$ Note that $$ \frac{\partial y_1}{\partial x_1}=2x_1=\frac{\partial y_2}{\partial x_2}\,,\quad \frac{\partial y_1}{\partial x_2}=-2x_2=-\frac{\partial y_2}{\partial x_1}\,. $$

  • $\begingroup$ Thanks for the help. However, I think we cannot use chain rule directly. We have to use whether the total derivative : $Df(g(x)) = D(f(g(x))[D(g(x))]$ or the jacobian matrix (I eventually solved this exercices with this method) : $Df(g(x)) = Jf.Jg$ $\endgroup$ – Mathieu Apr 17 at 9:11

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