Question

What is the equation of the line tangent to `f(x)=x tan2x` at `x=pi/8`?

Answer 1

`y=(pi/2+1)(x-pi/8)+pi/8`

Explanation:

Product Rule:
`f'(x) =(1) (tan(2x))+(x) (d/dx[tan(2x)])`
`f'(x) =tan(2x)+(x) [(2sec^2(2x)]`
`f'(x) =tan(2x)+2x sec^2(x)`

Evaluate the slope when `x=pi/8`
`f'(pi/8)=tan(2(pi/8))+2(pi/8)sec^2(2(pi/8))`
`m=1+pi/2`

Evaluate the function when `x=pi/8`
`f(x)=xtan(2x)=(pi/8)tan(2(pi/8))=pi/8`

Putting it all together:
`y-y_1=m(x-x_1)`
`y-pi/8=(pi/2+1)(x-pi/8)`
`y=(pi/2+1)(x-pi/8)+pi/8`