Question

Help me solve f(x)=7arctan(2sin(3x)), f′(x)=?

Based on what I did it should be [42cos3x]/[3-2cos(6x)] but that's wrong.

Answer 1

`(42cos(3x))/(1+4(sin^2(3x))`

Explanation:

You'll have to memorize this rule:
`d/dx[arctan(u)]=(u')/(1+u^2)`
Use chain rule to find:
u'=`6cos(3x)`
Plug that and into the rule and remember to multiply by that pesky seven!