# Jacobian of a chained function

Lets say that I have the following function: $$y = (f \circ g \circ h)(x) = f(g(h(x)))$$

$$f:\mathbb{R}^{k} → \mathbb{R}, g : \mathbb{R}^{m} \to \mathbb{R}^k, h: \mathbb{R}^{n} \to \mathbb{R}^m$$

what is the dimension of the Jacobian matrix $$D(f \circ g \circ h)(x)$$?

Would the dimension be $$1\times k$$, since the function is only in function of one variable $$x$$?

• Let $f\circ g\circ h=F:\mathbb{R}^n\to\mathbb{R}$ (a scalar field.) What do you know about the Jacobian of scalar fields? Apr 29 '21 at 14:52

By the Chain Rule you have $$D(f\circ g\circ h)=Df\cdot Dg\cdot Dh,$$ where these last derivatives have:

$$Df$$ a $$1\times k$$-matrix;

$$Dg$$ a $$k\times m$$-matrix

and

$$Dh$$ an $$m\times n$$-matrix

which can be multiplied, then $$D(f\circ g\circ h)$$ is a $$1\times n$$-matrix.

That multiplication is of matrices which have functions in their entries.

Further, the evaluated version is $$D(f\circ g\circ h)|_p=Df|_{(g\circ h)(p)}\cdot Dg|_{h(p)}\cdot Dh|_p,$$ which is a product of matrices with numbers as entries, however the same analysis of the number of rows and columns applies, as above.

• thanx man, glad to be an "aider" Apr 29 '21 at 17:24

The composition $$f\circ g\circ h$$ is such that $$\underbrace{\mathbb R^n\overset{h}\to\mathbb R^m\overset{g}\to\mathbb R^k\overset f\to\mathbb R}_{\tilde f:\mathbb R^n\to\mathbb R}$$ so this shows that the Jacobian $$J$$ of the map $$\tilde f:=f\circ g\circ h:\mathbb R^n\to\mathbb R$$ is an element of $$\operatorname{M}(1\times n,\mathbb R)$$.

If (assuming a certain regularity of the function), for example, you consider smooth functions, you can think the Jacobian evaluated in a point $$p\in\mathbb R^n$$ as the representative matrix of the differential (linear) map, so in our case $$\tilde f_{p,*}:T_p\mathbb R^n\to T_{\tilde f(p)}\mathbb R$$ respect to the standard basis of the tangent spaces $$T_p\mathbb R^n$$ and $$T_{\tilde f(p)}\mathbb R$$, so respectively $$\bigg\{ \dfrac{\partial}{\partial x^i}\bigg |_{p}\bigg \}$$ and $$\dfrac{d}{dt}\bigg|_{\tilde f(p)}$$. The Jacobian can be computed with the chain rule: $$J_{\tilde f}(p)=\underbrace{J_f(g(h(p)))}_{\in\operatorname{M}(1\times k,\mathbb R)}\cdot \underbrace{J_g(h(p))}_{\in\operatorname{M}(k\times m,\mathbb R)}\cdot \underbrace{J_h(p)}_{\in\operatorname{M}(m\times n,\mathbb R)}$$