Hello friends. So, welcome to lecture series
on multivariable calculus. So, in the last lecture we have seen what is mean by a partial
derivative of multivariable functions ok. So now, we will see second or higher order
partial derivatives. So, what do you mean by this let us see. So, again suppose z equal
to f (x y). Where x and y are independent variables and
z is a dependent variable. Now, we have seen we have already seen what do you mean by f(x)
at a point say a comma b it is simply del f upon del x, which is limit h tending to
0 f of a plus h comma b minus f of a, b upon h. And similarly, f y at a comma b is equals
to limit h tending to 0 f of a comma b plus h minus f of a, b upon h. Now, suppose we define del square of a point
del x square. Similarly, as we define second order derivative for single variable functions,
f double dash x here, we define second order partial derivative of f respect to x, it is
also denoted by f x x. So, at a point suppose we are defining it at a point a comma b, at
a point a comma b it is simply limit h tending to 0; f x of a plus h comma b minus f x of
a comma b upon h because, what does it mean basically? It means del f x upon del x or
at a comma b it means del by del x of f(x); that means, now you have to take f(x) a new
function and you have to apply the definition of del upon del x in f x. So, we have applied
definition of del by del x in f(x) as f (x a plus h) minus b; f x of a comma b minus
upon h, we simply replace f by f x in this definition so that we can get del square f
upon del x square. Now, similarly if we define in del square
f upon del y square, that is second order partial derivative of f respect to y at say
a comma b, which is can be defined as del by del y of del f upon del y at a comma b
or also denoted by f yy at a comma b is simply given by limit h tending to 0; f of f y of
because now, you have to apply the definition of del of del upon del y on f y ok, and del
upon and f definition of f y is this. So, you have to simply replace f by f y here so;
that means, f y of a comma b plus h minus f y of a comma b upon h. So, this is how we can define second order
partial derivative of f respect to x or respect to y. Now, come to mix order partial derivatives;
mixed second order partial derivative of f respect to x or y. Now, suppose you want to
define del square f upon del x del y at say a comma b; it is del by del x of del f upon
del y at a comma b or we can say, it is f del of f y of del f upon del x at a comma
b. So, this is by definition we have to replace
f by fy here, within this definition because, it is del f upon del x and instead of f we
are now having fy. So, that will be limit h tending to 0 fy of a plus h comma b minus
fy of a comma b upon h. Now, this mixed order second order mixed partial
derivative can also be represented it as f of y x because, it is what? It is fy respect
to x; that means, del upon del x of f y; that means, I differentiate partially this function
respect to x, that is del upon del x of fy which is same, which is same which is here.
So, we can also represent this partial derivative as fyx. Now, the second one is del square f upon del
by del x, which is at a comma b which is also represented as del by del y of del f upon
del x at say a comma b or it is f of x y at a comma b. It is by the definition we can write it like
this, it is limit x tending to 0. Now, you have to apply the definition of del upon del
y on fx. So, it is fx at a comma b plus h minus f of fx at a comma b upon h. So, this
is how, we can define fxy or fyx. So, fxy or fyx are also called mixed second
order partial derivative. Now, let us try some problems based on this. So, these are
all second order partial derivatives respect to x or y. Now, suppose f is 2 x minus 3 y whole cube.
Now, what is fx? it is; t cube is 3 t square and del by del x of t again which is 2 here,
and 2 into 3 is 6 a 6 times 2 x minus 3 y whole square, and again and fy is similarly
it is 3 t square t is 2 x minus 3 y and it is del by del y of t again, which is minus
3 into 3 is minus 9; 2 x minus 3 y whole square. Now, suppose you want to compute fxx; means,
del by del x of f x. Now, del by del x of fx is; you have to differentiate
fx respect to x again partially. So, when you differentiate this partial derivative
respect to x again. So, what we will obtain this is again 6 into t square, which is 6
into 2 2 into 2 x minus 3 y this for 1 into del by del x of t again, which is 2, 6 into
4 is 24, 24 2 x minus 3 y. Now, if you want to compute fyy, which is del by del y of fy
and it is differentiate this function respect partial respect to y. So, it is minus 18 2 x minus 3 y and del by
del y of t again. So, that will be 54 times 2 x minus 3 y ok. Now, suppose you want to
compute fxy, which is fx of y ok; that means, del by del y of fx ok. So, del by del y of
fx this is fx del by del y of this function, that is differentiate this partial respect
to y and; that means, it is 12 2 x minus 3 y del by del y of this again respect to y,
that is minus 3 into 12 is minus 36 times 2 x minus 3 y. Now, fyx suppose fyx; fy respect to x, that
is del by del x of f y; that means, it is f y you have to differentiate this partial
respect to x; that means, minus 18 2 x minus 3 y del by del y of or del by del x of this
function 2 x minus 3 y, which is 30 minus 36 times 2 x minus 3 y. So, that is how we
can compute second order partial derivatives respect to x or y ok. Now, see this again
say second problem. So, what is second problem now, it is f equal
to x upon x square plus y square and x y should not equal to 0 0 because, there is a problem
in 0 0. So, we are excluding 0 0 from the domain. Now, what is fx? First you compute
fx. Now, x is a numerator x is a denominator also. So, we have to apply quotient rule,
that is denominator derivative of numerator minus numerator derivative of denominator,
that is 2 x upon denominator whole square. So, this will be equals to y square minus
x square upon x square plus y square whole square ok. Now, what is f y? Now, it is x
constant for this partial derivative of f is respect to y. So, we take x into del by
del y of 1 upon x square plus y square. So, it is x into 1 by t, that is minus 1 by t
square. So, it is minus 1 upon t square then t again, that is del by del y of t and this
is 2 x y upon x square plus y square whole square and with negative sign. So, this is
del f upon del y. Now, you want to compute second order partial
derivatives again we can use this thing fxx, which is del by del x of fx we have to differentiate
this again respect to x ok. So, it is; denominator as it is derivative of numerator, which is
minus 2 x minus numerator derivative of denominator it is 2 times x square plus y square and it
again upon denominator whole square ok. So, this will be minus 2 x x square plus y
square and it is x square y square 1 is cancelled out it is minus 2 x of square y square and
it is minus 4 x y square minus x square upon x square plus y square whole cube ok, and
it is it is equals to it is minus 2 x cube it is plus 4 x cube. So, it is 2 x cube and
it is minus 2 x y square, it is minus 4 x that is minus 6 x y square upon x square plus
y square whole square whole cube. Similarly, we can find fyy also. Now, let us compute one more term here, it
is one of the mixed derivative anyone say we compute fxy that is, del by del y of fx
that is simply means, you have to differentiate this partially respect to y. So, again denominator
as it is derivative of numerator minus numerator derivative of denominator 2 t t again that
is 2 y upon denominator whole square. So, this is further equal to cancel a, so square
y square from numerator and denominator. So, we will obtain 2 y into x square plus
y square and it is minus 4 y y square minus x square upon x square plus y square whole
cube. So, this is 2 y cube minus 4 y cube, that is minus 2 y cube and it is 2 y x square
it is minus plus 4 y x square, that is plus 6 y x square upon x square plus y square whole
cube. And similarly, we can find fyx. So, that is how we can find f xx f yy f xy or
f yx. Now, let us solve one problem of 3 variables.
Now, suppose f(x y z) is; x raised to power x, y raised to power y, z raised to powers
z. x y and z are non-negative. Now, you have to find all second order partial
derivatives for this function. So, how will you compute fx. So, you have to take log first.
So, log of f will be equals to x log x plus y log y plus z log z. Now, differentiate this
equation partially respect to x. So, log t that is 1 upon t and t again, that is fx.
Now here, both are function of x. So, you apply product rule; first as it is derivative
second plus second as it is derivative of first and these two are independent of x.
So, derivative will be 0 partial derivative will be 0. So, this implies f x will be f into 1 plus
log x of course, it is base e. Now, similarly f
y will be f into 1 plus log y and f z will be f into 1 plus log z. Now, suppose you want
to compute f xx. So, how can you compute this? It is del by del x of fx which is so, so you
have to differentiate this again respect to x. So, this is also a function of x and this
is also a function of x. So, you have to apply product rule. So, f first as it is, derivative
of second plus second as it is, derivative of first fx and what is fx? what is fx? fx
is this. So, it is f into 1 plus log x. So, this is
f xx. Now, similarly f yy will be because, it is symmetrical in x, y and z. So, we can
simply write f yy as f upon y plus 1 plus log y whole square; f zz as f upon z plus
1 plus log z whole square ok. Now, suppose you want to find out f suppose f xz that is
f x at z, that is del by del z of f x. Now, what is f x and you have to differentiate
this f x respect to z partially. Now, this is independent of z because, x y z are independent
of each other independent variables ok. So, it is 1 plus log x you can take outside
and this is the function of x y z. So, partial derivative of respect to z will be f z and
f z is f plus 1 plus log z. So, it is f 1 plus log z log x into 1 plus log z. So, again
using a symmetry we can simply say that, f yz will be f into 1 plus log y into 1 plus
log z and similarly, f xy will be f 1 plus log x into 1 plus log y ok. So, that is how
we can find out all other all mixed second order partial derivative for this function.
Now, the next problem proved at the function this satisfy this equation. So, again this
is a simple problem let us see how can we. So, show this what is f here? F is e raised
to power 3 x plus 4 y and cos 5 z ok. Now, what is f x now, it is e raised to power t.
So, e raised to power t is e raised to power t derivative cos 5 z remain as it is and t
again that is del by del x of t again that is 3 x plus 4 y, which is 3 times e raised
to power 3 x plus 4 y into cos 5 z ok. Now, what is f xx? the derivative of this again
partial derivative of this again respect to x. So, that will be 9 e raised to power 3
x plus 4 y cos 5 z this is one. Now, compute f y; f y is 4 e raised to power
3 x plus 4 y into cos 5 z we can directly see, derivative of e raised to power t is
e raised to power t and t again, which is 4 cos 5 z remain as it is. So, it is 4 e raised
to power 3 x plus 4 y into cos 5 z. Now, this implies f yy will be now, differentiate this
again partially respect to y. So, what we would obtain this is 16 e raised to power
3 x plus 4 y cos 5 z. So, this is second equation. Now, what is
f z? f z is now, this is free from z remain as it is cos t is minus sin t and t again,
which is 5 and this implies f zz will be now, differentiate this again partially respect
to z. So, what we will obtain this is 5 into 5 is 20 minus 25 it is e raised to power 3
x plus 4 y into cos 5 z because, derivative of sin is cos. So, this is third equation. Now, when you
add when you add 1 2 and 3. So, what we would obtain we will obtain f xx plus f yy plus
f zz in the left-hand side and the right-hand side this is same and 16 plus 9 is 25 minus
25 is 0. So, this is equal to 0. So, hence we have proved. So, this equation is basically
called Laplace equation, this equation we call as Laplace equation. So, we can say that
this function e raised to power 3 x plus 4 y into cos 5 z satisfy a Laplace equation
now, what can we say about mixed second order partial derivatives, that is f xy or f yx
are they always same or there should be some condition on function f. So, that we can say,
that they are equal at some point. So, let us discuss this example; example is
f x y is equals to xy x square minus y square upon x square plus y square x y not equal
to 0 0 and 0 when x y is equal to 0 0 ok, let us discuss this example. So, things will
be clear after setting this example. So, let us compute f xy at origin and f yx at origin
for this particular problem ok. So, what is f xy at origin? It is f x at origin
is f x at y at origin, that is del by del y of f x at origin and that is limit h tending
to 0. So, what it is it is limit h tending to 0 f x at 0 comma 0 plus h minus f x at
0 comma 0 upon h. So, this is f xy at 0 comma 0. Now similarly, what is f yx at 0 comma
0 it is f y at x 0 comma 0, which is del by del x of f y at 0 comma 0; that means, it
is limit h tending to 0 f y at 0 plus h comma 0 minus f y at 0 comma 0 whole divided by
h. So, first we will compute f x at 0 comma h
f x at 0 comma 0 and f y at this then, we can compute f x y or f y x 0 comma 0. So,
what is f x at 0 comma 0 it is limit h tending 0 f at 0 plus h comma 0 minus f at 0 comma
0 upon h ok. So, it is now, you take x as h y as 0 when you take x as h y as 0 y as
0. So, this will be simply 0 and f 0 0 is 0. So, 0 minus 0 is 0. So, this value is simply
0. Now, you compute f x at 0 comma h we need now, we need f x at 0 comma h ok. Now, this
is say now take limit k tending to 0 because, h is already here. So, it is f at 0 comma k h minus f at f at
0 comma h upon k ok, at 0 comma h. Now, you replace x by k and y by h. So, what we will
obtain limit k tending to 0 it is h k into k square minus h square upon k square plus
h square minus 0 comma h is when you substitute x as 0 y as h. So, it is 0 whole divided by
k. So, this k will cancel with this k and when you take k tending to 0 here. So, it will be, it will be minus when you
take a tend to 0 it is minus h cube upon h square, which is minus h. So, what we can
say about f xy at 0 comma 0 now you can substitute it over here, f x at 0 comma h is minus h
you can substitute this value as minus h f x at 0 comma 0 is 0. So, it is 0 upon h. So, this expression will be
limit h tending to 0 minus h minus 0 upon h, which is minus 1. So, so we can say, that
this value at 0 comma 0 is minus 1. Now similarly, we try to compute f yx at 0 comma 0. Now,
what is f y at 0 comma 0 it is limit h tending to 0 f at 0 comma 0 plus h minus f 0 0 upon
h now, again when you take x as 0 y as h ok. So, it will be 0 and f 0 0 is 0. So, this
value is 0. Now, when you take f y at say 0 comma h, h comma 0 you see, we need f y
at h comma 0. So, f y at h comma 0 is limit k tending to
0 f at we need f y; that means, we have to change y only x has remained fixed 0 comma
h comma 0 plus k minus f at h comma 0 divided by k. Now, this is limit k tending to 0 now
here, you replace x by h and y by k. So, it is h k. So, it is h k into h square minus
k square upon h square plus k square this minus f at h comma 0 x is h y is 0 when you
take y as 0. So, this expression is 0. So, it is 0 upon k. So, this k will cancel with
this k and when you take k tending to 0. So, k will be 0. So, it is h cube upon h square
which is h. So, what can we say about f yx? So, this will
be equal to now here limit h tending to 0 what is f y at h comma 0 it is h. So, it is
h minus f y at 0 comma 0 is 0. So, it is 0 upon h. So, this value is 1. So, what do we
conclude from here. So, we have concluded that, f yx at some point may not be equal
to f xy it is 1 it is minus 1. For some function f xy may not be equal to f yx. So, what is the additional condition required
on f. So, that we can say, that the mixed second order partial derivatives are equal
ok. So, that condition is basically we have Euler’s theorem, which states that if the
function f xy and it is partial derivatives f x f y f xy and f yx are defined throughout
an open region containing a point a comma b and all are continuous at a comma b then,
only we can say that f xy is equals to f yx otherwise they may not be equal ok. So, this is all about second or partial derivatives.
Now, we can define higher order partial derivative also like, suppose z equal to f xy the function
of x or y. So, we can define say 4th order partial derivative also with respective to
x, which is f xxxx we can also define say del cube f upon del x square del y, which
is simply del square upon del x square of del f upon del y or del by del x of del square
f upon del x del y ; that means, del by del x of f it is y x or del y x of x.
So, these are all representation of the same expression basically similarly, we may have
any higher derivatives suppose del cube f upon del y cube, which is f yyy and so on.
So, similarly we can compute these values also for any function f so. Thank you very much.
