I have the following function:

$f(x, y, z) = (x^2 + \ln y, \sqrt{y − z}).$

I have found that the partial derivatives are continuous when $y> z$ and $y> 0$

The question is:

How to calculate the Jacobian matrix when the function is differentiable?

  • $\begingroup$ Do you know the definition of the Jacobian matrix? $\endgroup$ Nov 7 '17 at 10:15
  • $\begingroup$ I see here three variables! $\endgroup$
    – Nosrati
    Nov 7 '17 at 10:31

Let $f:ℝ^n→ℝ^m$ be a function with arguments $(x_1,…,x_n)$ and components $(f_1,…f_m)$. The Jacobian is defined as the $m×n$ system: $$J(f) = \left(\frac{∂f_i}{∂x_j}\right)_{i,j=1}^{m,n}=\pmatrix{\frac{∂f_1}{∂x_1} & \frac{∂f_1}{∂x_2} & \dots & \frac{∂f_1}{∂x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{∂f_m}{∂x_1}& \frac{∂f_m}{∂x_2} & \dots &\frac{∂f_m}{∂x_n}} $$

In your case it is $$f(x, y, z) = \pmatrix{x^2 + \ln(y) \\ \sqrt{y − z}}=: \pmatrix{f_1(x,y,z)\\f_2(x,y,z)}. $$

So the Jacobian is a $2×3$ matrix

$$J(f) = \pmatrix{2x& \star & \star\\ 0 & \star & \star}$$

Can you take it from here?


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