find nth derivative of y=x^3. e^x.cosx
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1
Answer:
Let y=e
6x
cos3x
dx
dy
=
dx
d
(e
6x
.cos3x)=cos3x.
dx
d
(e
6x
)+e
6x
.
dx
d
(cos3x)
=cos3x.e
6x
dx
d
(6x)+e
6x
.(−sin3x).
dx
d
(3x)
=6e
6x
cos3x−3e
6x
sin3x ....(1)
∴
dy
2
d
2
x
=
dx
d
(6e
6x
cos3x−3e
6x
sin3x)=6
dx
d
(e
6x
cos3x)−3
dx
d
(e
6x
sin3x)
=6[6e
6x
cos3x−3e
6x
sin3x]−3[sin3x
dx
d
(e
6x
)+e
6x
dx
d
(sin3x)
=36e
6x
cos3x−18e
6x
sin3x−3[sin3x.e
6x
.6+e
6x
cos3x.3]
=27e
6x
cos3x−36e
6x
sin3x=9e
6x
(3cos3x−4sin3x)
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