# Thread: chain rule (jacobian matrix)

1. ## chain rule (jacobian matrix)

$z = sin (u) cos (v); u=3x^2, v=x-3y$

$\left(\begin{array}{cc}cos(u)cos(v)&-sin(v)sin(u)\end{array}\right) \left( \begin{array}{cc} 6x&0\\1&-3\end{array}\right)$ $=\left(\begin{array}{cc}6xcos(3x^2)cos(x-3y)-sin(x-3y)sin(3x^2)\\0-3sin(x-3y)sin(3x^2)\end{array}\right)$

I have a feeling this is incorrect

2. ## Re: chain rule (jacobian matrix)

It is incorrect, but very close. You want the transpose of the final matrix (it should be two columns instead of two rows).

3. ## Re: chain rule (jacobian matrix)

$\left( \begin{array}{cc}6xcos(3x^2)(x-3y) - sin(x-3y)sin(3x^2)&3sin(x-3y)sin(3x^2)\end{array}\right)$

thanks, got the matrix multiplication mixed up