So now, we've got our a bit complicated definition here.
Let's see what it actually means
for our some given functions.
For example, let us revisit the definition once again.
I'm going to stress something out for you.
Firstly, we all need to understand that
basically we can enroll in two possible manners here.
We can either substitute or fix
all the variables except
the variable of differentiation for example,
we need to substitute y with b and then find
the derivative by the definition or any other technique.
Here comes the difference.
One can differentiate our function
at every point and then
substitute our variable x
with a point we're actual interested in,
or just find a derivative at this very point.
It's not a big difference or every thing will be true.
So let us see what it actually means.
For example, let us take on
quite simple extra function here,
square root from absolute value of x product y.
Assume that we're in need of finding
partial derivatives at zero point.
So what we're going to do?
We are going to consider
function which is a restriction of
powers of four y equals to zero.
So as a result,
we get function F to root x and zero which is
square root from x
multiplied by zero which is simply zero,
for all possible axis right?
So the derivative of zero it's pretty obvious that it is
zero and as a result
it is zero if we substitute x with zero.
Same thing applies obviously for the derivative.
Partial derivative to [inaudible].
Why is so we've actually find both partial
derivative at a given point right? That was easy.
So maybe assume that we need to
find the derivative at some arbitrary point right?
In order to do so first of all,
let us look at this nice surface here.
That's kind of what we've extracted right,
because we've worked at
our partial derivatives are equal to zero.
So we get horizontal tangent lines
of four hour zero zero point.
So as for arbitrary point,
let us cut them as
a more complicated but actually more nice function
here, more smooth function.
Here x squared plus y squared multiply it by sum of
product x and y and we are going to
calculate our derivative at pretty much any point.
Here is what I'm going to do,
I'm going to just assume that we're going to substitute.
For example y with a b just on the [inaudible].
So what we're gonna do we're gonna write
our function that x squared plus b
squared multiplied by sine of bx and
right here we are going to just
find the derivative combined x.
We're looking at the derivative of the project
and thus we are going to roll as usual.
We need to calculate
the derivative of the first one which
is x times two and multiply it by the second,
and then add vice-versa.
The first one, the first function x squared
plus b squared multiplied by the derivative of
sine of bx which is essentially a composition of
two functions of sine with cosine
as a derivative and bx which is quite easy.
The derivative of it's equals to b.
So we find our
partial derivative is on the sine will have to
do is substitute our X
with points a and del give us our answer.
Sometimes people usually omit
the step of substituting y with b and to x plus y.
Because obviously we are talking
about the values of this partial derivative at any point.
At point wherever a and b can be
[inaudible] from the real plane.
So we just can't skip our substitution with
a and b and just roll with
the derivative in terms of x and y.
That's always true, but
I'm just stressing you out but that you
always think about fixating
your variables in the first place.
So now we know how to calculate
our derivatives with substitution
and that was quite easy.