I'm experiencing some troubles with an exercise that a friend asked me. Hence we decided to ask it on here to get some more explanations!

This is the exercise: find the Jacobian of the following function, at the point (1, 0, 0):

$$f(x, y, z) = g(3x^5 + y^3 + \phi(x, y, z),\ \ln(1+x^2y^4) - \eta(x, y, z))$$


$g\in C^1(\mathbb{R}^2, \mathbb{R}^3)$

$\eta, \phi \in C^1(\mathbb{R}^3, \mathbb{R})$

$\phi(1, 0, 0) = 1$

$\eta(1, 0, 0) = -1$

$\nabla \phi(1, 0, 0) = (1, 1, 1)$

$\nabla\eta (1, 0, 0) = (-1, -1, -1)$

$$Jg(4, 1) = \begin{pmatrix} e & \pi \\ -\pi^2 & e \\ 3 & 5 \end{pmatrix} $$

I think that what confuses me the most is the $f$ as $f = g(...)$

Shall we use the chain rule or what?

I found that the Jacobian for a composite function, let's say for $f\circ g$ is given by

$$J(f\circ g)(a) = \begin{pmatrix} \frac{\partial f_1}{\partial g_1}\frac{\partial g_1}{\partial x} + \frac{\partial f_1}{\partial g_2}\frac{\partial g_2}{\partial x} + \frac{\partial f_1}{\partial g_3}\frac{\partial g_3}{\partial x} & \frac{\partial f_1}{\partial g_1}\frac{\partial g_1}{\partial y} + \frac{\partial f_1}{\partial g_2}\frac{\partial g_2}{\partial y} + \ldots & \frac{\partial f_1}{\partial g_1}\frac{\partial g_1}{\partial z} + \frac{\partial f_1}{\partial g_2}\frac{\partial g_2}{\partial z} + \ldots \\ \frac{\partial f_2}{\partial g_1}\frac{\partial g_1}{\partial x} + \frac{\partial f_2}{\partial g_2}\frac{\partial g_2}{\partial x} + \ldots & \frac{\partial f_2}{\partial g_1}\frac{\partial g_1}{\partial y} + \ldots & \frac{\partial f_2}{\partial g_1}\frac{\partial g_1}{\partial z} + \ldots \end{pmatrix} $$

What I don't understand is what $g_1$ and $g_2$ and $g_3$ are.

  • $\begingroup$ Yes, as the function is composite, you have to apply the chain rule. $\endgroup$ – LutzL Jan 15 '17 at 17:54

$g_i$ are the components of $g$, as we have $g = (g_1,g_2,g_3)$. But I wouldn't use that horrible looking formula. Instead, $f = g \circ h$, where $h = (h_1,h_2): \Bbb R^3 \to \Bbb R^2$, so applying the chain rule, $Jf(a) = Jg(h(a)) Jh(a)$. We have:

$$Jg(b) = \begin{bmatrix} \frac{\partial g_1}{\partial x}(b) & \frac{\partial g_1}{\partial y}(b) \\ \frac{\partial g_2}{\partial x}(b) & \frac{\partial g_2}{\partial y}(b) \\ \frac{\partial g_3}{\partial x}(b) & \frac{\partial g_3}{\partial y}(b) \end{bmatrix}$$

$$Jh(a) = \begin{bmatrix} \frac{\partial h_1}{\partial x} (a) & \frac{\partial h_1}{\partial y} (a) & \frac{\partial h_1}{\partial z} (a) \\ \frac{\partial h_2}{\partial x} (a) & \frac{\partial h_2}{\partial y} (a) & \frac{\partial h_2}{\partial z} (a) \end{bmatrix}$$

Now compute and substitute $a=(1,0,0)$ and $b = h(1,0,0)=(4,1)$.

  • $\begingroup$ That is really enlightening! Thank you! $\endgroup$ – Von Neumann Jan 15 '17 at 19:32
  • $\begingroup$ @AlanTuring you're welcome! $\endgroup$ – user384138 Jan 15 '17 at 19:33

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