If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Computing a Jacobian matrix

## Video transcript

so just as a reminder of where we are we've got this very nonlinear transformation and we showed that if you zoom in on a specific point while that transformation is happening it looks a lot like something linear every reason that you can figure out what linear transformation that looks like by taking the partial derivatives of your given function the one that I defined up here and then turning that into a matrix and what I want to do here is basically just finish up what I was talking about by computing all of those partial derivatives so first of all let me just rewrite the function back on the screen so we have it in a convenient place to look at the first component is X plus sine of Y sine of Y and then Y plus sine of X was the second component so what I want to do here is just compute all of those partial derivatives to show what kind of thing this looks like so let's go ahead and get rid of this word then I'll go ahead and kind of redraw the matrix here so for that upper left component we're taking the partial derivative with respect to X of the first component so we look up at this first component and the partial derivative with respect to X is just 1 since there's 1 times X plus something that has nothing to do with X and then below that we take the partial derivative of the second component with respect to X down here and that guy the Y well that looks like a constant so nothing happens and the derivative of sine of X becomes cosine of X and then up here we're taking the partial derivative with respect to Y of the first component that upper one here and for that you know partial derivative of X with respect to Y is 0 and partial derivative of sine of Y with respect to Y is cosine of Y and then finally the partial derivative of the second component with respect to Y looks like 1 because it's just 1 times y plus some constant and this is the general Jacobian as a function of x and y but if we want to understand what happens around the specific point that started off at well I think I recorded here at negative 2 1 we plug that into each one of these values so when we plug in negative 2 1 so go ahead and just kind of again rewrite it to remember we're plugging in negative 2 1 as our specific point that matrix as a function kind of a matrix valued function becomes one the next we have cosine but we're plugging in negative 2 for X cosine of negative 2 and if you're curious that is approximately equal to I calculated this earlier negative 0.42 if you just want to think in terms of a number there then for the upper right we have cosine again but now we're plugging in the value for Y which is 1 and cosine of 1 is approximately equal to 0.5 4 and then bottom right that's just another constant 1 so that is the matrix just as a matrix full of numbers and just as kind of a gut check we can take a look at the linear transformation this was supposed to look like and notice how the first basis vector the thing it got turned into which is this vector here does look like it has coordinates 1 and negative 0.42 right it's got this rightward component that's about as long as the vector itself started and then this downward component which I think that's you know pretty believable that that's negative 0.42 and then likewise this second column is telling us what happened to that second basis vector which is the one that looks like this and again it's it's Y component is about as long as how it started right a length of 1 and then the rightward component is around 1/2 of that and we actually see that in the diagram but this is something you compute again it's pretty straightforward you just take all of the possible partial derivatives and you organize them into a grid like this so with that I'll see you guys next video