find the second order derivative of tan^-1 3x
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Answered by
1
Answer:
kwkowkdkjddjjdksekkskakallppPassssed
Answered by
1
Answer:
I hope it will help you
Step-by-step explanation:
Let y=e
6x
cos3x. Then,
dx
dy
=
dx
d
(e
6x
)cos3x+e
6x
dx
d
(cos3x)
⇒
dx
dy
=6e
6x
cos3x−3e
6x
sin3x
⇒
dx
dy
=3e
6x
(2cos3x−sin3x)
⇒
dx
2
d
2
y
=3
dx
d
(e
6x
)(2cos3x−sin3x)+3e
6x
dx
d
(2cos3x−sin3x)
⇒
dx
2
d
2
y
=18e
6x
(2cos3x−sin3x)+3e
6x
(−6sin3x−3cos3x)
⇒
dx
2
d
2
y
=9e
6x
{4cos3x−2sin3x−2sin3x−cos3x}=9e
6x
(3cos3x−4sin3x)
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