Question

What is the equation of the line tangent to `f(x)=x sin^2(2x)` at `x=pi/12`?

Answer 1

`y=(sqrt(3)/123*pi+1/4)*x-1/144*pi^2*sqrt(3)`

Explanation:

We have
`f(x)=xsin(2x)^2`
then
`f'(x)=sin^2(2x)+x*2*sin(2x)*cos(2x)*2`
`f'(x)=sin^2(2x)+2x+cos(2x)`
we get
`f'(pi/12)sin^2(pi/6)+2*pi/12*cos(pi/6)`
`f'(pi/12)=sqrt(3)/12*pi+1/4`
and
`f(pi/12)=pi/48`
and the Tangent line is given by
`y=(sqrt(3)/12*pi+1/4)x+n`
where
`n=-1/144*pi^2*sqrt(3)`