Hi Friends, So in this video we are gonna
see nth derivative of some more
functions now let us take an example if
y is a x plus b raised to minus m then
what is the nth derivative of this now
to find the nth derivative we are going
to follow one approach that is we will
find the first derivative that is y1
second derivative y2 then third
derivative y3 and from this first three
derivatives we will predict what could
be the y n so let's start with this now
y1 as you all want to know it is nothing
but minus 2 raised to ax plus b raised
to minus 2 minus 1 this is by using the
formula of derivative into a now if you
will find the y2 from the answer of y1
it is nothing but minus m into minus 2
minus 1 ax plus b raised to minus 2
minus 2 into a square because already we
were having one a in y1 and one way of
this y2 so it becomes a square let us
find what could be y3 so y3 is nothing
but minus M minus M minus 1 minus M
minus 2 ax plus B raised to minus M
minus 3 into a cube and now it's time to
find out what could be the Y n now to
find y n we will rearrange all the terms
which we got in y1 y2 and y3 so if you
all will observe then we can say that in
y1 we are getting M as negative so here
the minus 1 raised to 1
is there now M is constant I will keep
it as it is a is constant I will keep it
as it is and here this term is nothing
but ax plus B raised to minus sign
outside M plus 1 similarly this second
term if will rewrite is from the first
term we are getting minus sign from the
second term we will get minus M common
so this will become minus 1 raised to 2
because minus sign is coming 2 times
next M will remain as it is it will
become positive this will become
positive that is M plus 1 this a square
will remain as it is and here we will
get ax plus B raised to by taking minus
and outside we will get M plus 2 in
bracket similarly we will rewrite the
third term now in the third term we will
get three negative signs one from this
minus M one from this minus a minus one
and one from minus M minus two so I will
rewrite this as minus 1 raised to 3
rewrite the terms that is m+ m plus 1
positive M plus 2 positive next a cube
as it is and this term we will rewrite
as ax plus B by taking minus n outside
we will get M plus 3 inside the bracket
now let's guess what could be the Y n so
as we were getting minus 1 raised to 1
in Y 1 minus 1 raised to 2 in the Y 2
minus 1 raised to 3 in Y 3 I will get
minus 1 raise to n in Y n similarly we
were getting M in case of Y 1 M into M
plus 1 in case of Y 2 M into M plus 1 M
plus 2 in Y 3 so in case of Wyatt we
will get M M plus 1 M plus 2 so on till
M plus
n minus one since in case of y3 the last
term was two in case of right to the
last term was one
hence in case of Y and the last term
will be n minus one next in case of y
one we were getting a raise to one in
case of y 2 we were getting a square in
case of Y three we were getting a cube
hence in case of Y n we will get a raise
to n similarly in case of Y 1 we are
getting ax plus B raised to M plus 1
minus n outside in case of Y 2 we are
getting minus of M plus 2 in case of Y 3
we are getting minus of M plus Li
similarly in case of Y and we will get
minus of M plus n so let's rewrite this
so this will become minus 1 raise to n M
into M plus 1 into M plus 2 so on till M
plus n minus 1 e raise to n now this
will terr this term will go in the
denominator and this will become a X
plus B raised to positive M plus n now
let's reduce the numerator so to reduce
the numerator what I am doing I am
converting this in terms of factorial
now let us observe volume here the last
term is M so what I will do is I will
further multiply this side with the
terms like 1 2 so on till M minus 1 so
here remaining terms as it is that is M
M plus 1 M plus 2 and so on till M plus
n minus 1 e raise to n now as I am
multiplying numerator with the terms
like 1 into 2 till n minus 1 I have to
multiply the denominator with the
similar term so in the denominator I
will get 1 into 2 into dot-dot-dot till
M minus 1 into a X plus B raised to M
plus n
now let's see what we got so in the
numerator minus 1 raise to n as it is
and here this 1 into 2 into dot dot dot
till M plus n minus 1 has become M plus
n minus 1 factorial next a raise to n as
it is in the denominator we got 1 into 2
into dot dot dot till M minus 1 so it is
again M minus 1 factorial and ax plus B
raised to M plus n will remain as it is
and this becomes the formula that is nth
derivative of ax plus B raised to minus
M now this formula is very useful to
find out the nth derivative of multiple
terms so let's see one more corollary
which is related with this so in that
corollary we will try to find out n the
derivative of y equal to ax plus B
raised to minus 1 so if you will compare
this with the derived formula or if
you'll compare this with our question
then you can say that M is equal to 1 so
now what I will do is in the DL formula
I will just substitute M as 1 so let's
see what we get
so we will get Y n as so minus 1 raise
to n will remain as it is as it don't
have M here when I am substituting M as
1 so 1 minus 1 is 0 plus n so only n
factorial will remain a raise to n as it
is because it doesn't contain M in the
denominator 1 minus 1 0 and we all know
that 0 factorial is 1 next X plus B
raised to n plus M M is 1 so this will
become n plus 1 and this becomes one
more formula for nth derivative of y
equal to ax plus B raised to minus 1 or
y equal to 1
upon a X plus B so these formulas are
very useful to find out the nth
derivative of algebraic functions but
we'll see in the later part
