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# Computing a Jacobian matrix

## Video udskrift

- [Teacher] So, just as a
reminder of where we are, we've got this very
non-linear 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 and we 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 sin of y. Sin of y and then y plus sin 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 and 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 one. Since there's one 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 sin 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, partial derivative of x,
with respect to y, is zero and partial derivative of sin 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 one because it's just one
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 this specific
point that started off at, well, I think I recorded
it here at negative two, one, we plug that in to each
one of these values. So, when we plug in negative two, one. So go ahead and just
kind of again, rewrite it to remember we're plugging
in negative two, one as our specific point,
that matrix as a function, kind of a matrix valued function, becomes one, and then next we have cosine, but we're plugging in negative two for x, cosine of negative two,
and if you're curious, that is approximately equal to, I calculated this earlier. Negative zero point four
two, 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 one and cosine of
one is approximately equal to zero point five four;
and then bottom right, that's just another constant: one. 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 factor, the thing it got turned into,
which is this factor here, does look like it has coordinates one and negative zero point four two, 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 pretty believable that that's negative zero point four two. And then, likewise, this
second column is telling us what happened to that second basis factor, which is the one that looks like this. And again, its y component
is about as long as how it started, right, the length of one. And then the rightward component
is around half 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.