# Total differential of a compound function with vector basis

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 $$\frac{\partial y_1}{\partial x_1}=2x_1=\frac{\partial y_2}{\partial x_2}\,,\quad \frac{\partial y_1}{\partial x_2}=-2x_2=-\frac{\partial y_2}{\partial x_1}\,.$$
• 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) : $Df(g(x)) = Jf.Jg$ – Mathieu Apr 17 at 9:11