1.

Find nth derivative for y=sin3xcos2x​

Answer»

Step-by-step explanation:

start with a generalized product RULE by LETTING u=cos(2x)

f(x)=x⋅u

f′(x)=x⋅u′+u

f′′(x)=x⋅u′′+u′+u′=x⋅u′′+2u′

f′′′(x)=x⋅u′′′+u′′+2u′′=x⋅u′′′+3u′′

f(4)(x)=x⋅u(4)+u′′′+3u′′′=x⋅u(4)+4u′′′

... f(N)(x)=x⋅u(n)+nu(n−1)

now for the fun part,

u=cos(2x)

u′=−2sin(2x)

u′′=−4cos(2x)

u′′′=8sin(2x)

u(4)=16cos(2x)

...

note that both ±cos(2x) and ±sin(2x) can be written in the form cos(2x+b) where b is a multiple of π2

... u(n)=2ncos(2x+b)=2n[cos(2x)cosb−sin(2x)sinb]


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