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from Ekeeda Hello friends in this video
we are going to start few easy problems
on derivative and we are going to take
such three problems so let us start so
the first problem we have is 5 divided
by DX if Y is equal to X raised to 10
plus 10 raised to X plus e raise to X so
we are going to find the derivative of
this function using direct formulas and
without using the first principle so let
us see how to use this fir glass we have
y is equal to X raised to 10 plus 10
raised to X plus e raise to X so
basically there are 3 different
functions and there is plus sign in
between them so the first function is a
pure algebraic function and this
function can be compared with derivative
of x raised to M because the base X
should be the variable and the power n
should be constant which can be seen in
the first part that is X raised to 10
the second part can be compared with
derivative of exponential function that
is e raise to X and the last part can be
similarly compared with the derivative
of again an exponential function e raise
to X friends the formula for this is n X
raise to n minus 1 second it is 2x log a
and the third one we have already seen
in the previous videos the derivative of
e raise to X is e raise to X now let us
start with the sum so we'll say
differentiate the above function
with respect to X so with respect to x
wi d stands for with respect to so the
derivative of Y with respect to X will
be dy by DX and on the right side we'll
write D by DX of X raised to 10 plus 10
raised to X plus e raise to X now this
derivative will be separated to each of
the functions so let us start with it so
this will become D by DX of X raised to
10 plus D by DX of 10 raise to X plus D
by DX of e raised to X now to find a
derivative of X raised to 10 we are
going to use this formula that is NX
raise to n minus 1 now here n the value
of n will be equal to 10 so that is
equal to n that is n then X raise to n
minus 1 means n minus 1 followed by +
derivative of n raise to X we have
derivative of e raised to X as e raise
to X log e so the derivative of 10
raised to X will be 10 raise to X log N
and followed by derivative of e raised
to X will be e raise to X so the final
answer can be written as dy by DX is
equal to 10x raised to 9 n minus 1 will
be equal to 9 plus 10 is 2x log n plus e
raise to X so this was the first problem
which we have solved using the direct
formulas let us start with the second
problem let us start with the second
problem so we have find dy by DX if Y is
equal to X plus 1 into X plus 2 so the
value of y is equal to X plus 1 into X
plus 2 to find a derivative of the
function we have two methods first we
can simplify this problem or
we can find the derivative directly
using the UV rule in the study we'll
simplify this function so y is equal to
simplification includes multiplication
of these two functions so we have X into
second term is X plus 2 again the second
term is plus 1 into X plus 2 so Y will
be equal to X into X will give you X
square plus 2 into X will give you 2x
plus 1 into X will give you 1 X and plus
1 into 2 will give you 2 after adding
the like terms that is 2x plus 1x which
will give you 3x plus 2 our function is
ready to differentiate so let us
differentiate this function with respect
to X so I would write differentiating
the above function with respect to X so
derivative of Y with respect to X will
give u dy by DX and here we have d by DX
of X square plus 3x plus 2 again the
next step will be followed by separating
the by DX to all the three terms so D by
DX of X square plus D by DX of 3x plus D
by DX of 2 in the next step we can take
the constants outside which are going to
be multiplied so d by DX of X square
plus we have 3 here as a constant so
this will become 3 into d by DX of X
plus we have 2 as a constant and we have
d by DX of 1 now let us see the next
step so the derivative of x square what
is the derivative of X square we can
compare it with naivety of x raise to n
derivative of X raise to n is n
X raise to n minus one so the derivative
of X square will become n that is 2 then
X that is X raise to n minus 1 that is X
raised to 1 then we have plus 3 into
derivative of X with respect to X is
equal to 1 followed by a constant so
derivative of a constant will be equal
to 0 so the final answer will be 2x plus
3 so this is the derivative of the given
function now let us start with the third
problem so in the third problem we have
if Y is equal to e raise to X into sine
X find dy by DX in this case we have two
functions right here is 2x into sine X
yes it is into sine X so whenever the
two functions are getting multiplied we
will use UV rule so what is UV rule it
says D by DX of u into V what is U and V
both are functions in terms of X
similarly in this type we have both the
functions in terms of X so derivative of
U V is given as u into derivative of V
Plus V into derivative of U it is also
read as u into DV by DX plus B into D u
by D X so let us arrange this form in
this formula so first we will say
differentiate with respect to X so the
derivative of Y with respect to X will
be equal to dy by DX that is equal to
you can consider the first part as U and
second part as V it means the value of U
will be equal to e raise to X and the
value of V will be equal to sine X so
let us arrange in this format so you as
it is d by DX of V what is V V is sine X
plus V as it is min sine X into
derivative of U that is d by DX of e
raised to X now let us start
differentiating this functions
so the next step will be dy by DX is
equal to e raise to X derivative of sine
X is given as cos x plus sine X into
derivative of e raise to X we have Y
raise to X the final answer can be
written by taking a raise to X as common
so here is 2x common we have cos x plus
sine X so this is the derivative of the
given function y is equal to e raise to
X sine X I hope you have understood this
video thank you for watching this video
stay tuned with Akira and subscribe
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