Hi Friends, So in this video we are gonna
see nth derivative of some more
functions so let's start with it so let
us consider a function y equal to cos of
ax plus b now to find out the
antiderivative we will use a technique
by using which we will find the nth
derivative and the technique is first
find out first derivative then find out
second derivative then find out third
derivative after that from for second
and third derivative guess what could be
the nth derivative so i am going to use
that procedure so let us find out the
first derivative so y1 is called as
first derivative of function Y so the
derivative of cos x is minus of sin X
that is minus of sine of a X plus B into
hey now let us find out second
derivative that is why to but if I find
out the Y - oya I will get Y 2 as again
cause then if I find out divide 3 from
this Y 2 I will get a sign from that
cause so it means in each and every
derivative I am getting sign then cos
then sign then cos that is alternative
sign in cause and hence it will become
very difficult to guess what could be
the Y n because we do not know the value
of N and hence I will use one more
method that is every time to find out D
for the derivatives I will force the
generalize the given term in terms of
cos only because our question is in
terms of cost that here the
trigonometric ratio is cos hence every
time I will convert the answer in terms
of cost and then I will find out the
next derivative so if I want to convert
this minus sign term in cost then we all
know that cos of Pi by 2 plus theta is
minus sine theta so you are in this case
theta is ax plus B so I will say it is
PI by 2 plus a X plus B so I am using
the formula cos of Pi by 2 plus theta as
minus sine theta and as a is a constant
I will write it down as it is and this
will become my new value of y1 now let
us find out why - from this new value of
y1 so y2 of this course is again - sine
of PI by 2 plus ax plus B into a square
so if I want to find the next derivative
y3 I have to convert this minus sign
into a cost term and then we'll proceed
so to convert this minus sign term into
cos term we will again use the formula
cos of Phi by 2 plus theta because cos
of Pi by 2 plus theta is minus sine
theta so you are the theta is PI by 2
plus ax plus B so I will write it down
as cos of Pi by 2 plus theta theta is PI
by 2 plus ax plus B and this time this
will become a square next I can rewrite
this as cos of 2 times pi by 2 plus ax
plus B into a square so now my friends I
got the value of y1 I got the value of
y2 and now it's time to guess the value
of yn so I will say in Y and the answer
will be definitely a cost term and in
the angle Part C R when it was y1 I was
getting 1 once PI by 2 plus ax plus B
when it was y2 that time I was getting
twice PI by 2 plus ax plus B similarly
when I am finding antiderivative I will
get ax plus B plus n times pi by 2 so
bracket is
a X plus B plus n times pi by 2 and
similarly here in y1 you are having a
raise to 1 y2 we were having a raise to
2 and in Y and we will get a raise to n
and this is the nth derivative of cos of
ax plus B which we will use to solve
many problems now let's see and the
derivative of some more functions now
let's say the function is y equal to ax
plus B raised to M so to find out n the
derivative we will use the same approach
so let's start with first derivative so
therefore the first derivative that is y
1 will be given as M into a X plus B
raised to M minus 1 into a now let us
find out the second derivative so second
derivative we will get it from first
derivative that is y1
so let's differentiate this y1 to get y2
now m and as they are constant I will
keep them as it is so M and a as it is
next let's find the derivative of this
it is M minus 1 into a X plus B raised
to M minus 2 into a so now a will become
a square now let us find out the third
derivative from this
so therefore y3 will be C to find out
the third with derivative I am keeping
em a square and M minus 1 as it is
because they are constant and I am just
finding the derivative of this term so
the derivative of ax plus B raised to M
minus 2 is M minus 2 into a X plus B
raised to M minus 3 into a so it means
we are getting a square into a as a cube
next M M minus 1 M minus 2 into a
X plus B raised to M minus 3 and guys
now it's time to predict what could be
the Y n now let's observe the pattern
now in y1 I was getting a raise to 1 in
y2 I got a square in y3 I got a cube so
I can say that in yn I will get yes it
is a raise to n similarly let's observe
the remaining terms in y1 I was just
getting em in y2
I got em into M minus 1 in Y 3m M minus
1 M minus 2 so guys if you observe the
last term of this pattern of this series
then every time I am getting the last
term as one less than that particular
derivative so when I was finding the
third derivative I got two in the last
term when I was finding the second
derivative I got one in the last term so
can I say that when I will find out the
nth derivative I will get last term as n
minus 1 so the pattern will be M into M
minus 1 M minus 2 and so on till M minus
n minus 1 as a last term now pay
attention I am keeping the bracket for
that n minus 1 because the minus sign is
outside now let's observe the last term
now here in y1 I was getting ax plus B
raised to M minus 1 Y - I was getting ax
plus B raised to M minus 2 in y3 ax plus
B raised to M minus 3 so minus 1 in case
of y1 minus 2 in case of y2 similarly in
case of YY n I will get minus n so hence
the last term is a X plus B raised to M
- n now let's rewrite this term so we
will get wired as a raise to n into M M
minus 1 dot dot dot M minus n plus 1
into a X plus B raised to M minus n so
this is the formula to find out the nth
derivative of a X plus B raised to M now
guys let's observe one more thing that
this formula is applicable when this M
is less than the n so I'll say so guys
let's observe one more thing
that this formula is applicable when
this M is greater than this n so it
means if M is greater than n then this
formula is applicable now let's see if m
and n are equal then what happens if M
is less than n so when M and n both are
equal then that time in this last term
the changes will be like this so why n
will become here is 2 n that term will
remain as it is here as M and n are
equal I am saying that m and n are equal
so this n will become M and M minus M
will become 0 and I will get only 1 so
the last term of this series will be 1
similarly here the power will become ax
plus B M minus M that is 0 now let's
rewrite this a raised to n as it is next
M into M minus 1 M minus 2 till 1 is
nothing but M factorial and ax plus B
raised to 0 is 1 so it means I am
getting wine as a raise to n into M
factorial now let's see one more case
the third case is nothing but
what will happen if value of M is less
than n now let's see guys so in this
case this wine will be zero now how now
since M is less than n so if I'll try to
find out before second third fourth
fifth derivative at one step I will get
the derivative as zero and once I will
get zero in any of the derivative all
the remaining derivatives or the further
derivatives will be zero and hence wine
will be zero in that case
