Question

What is the equation of the line tangent to `f(x)=x tan^2x` at `x=pi/4`?

Answer 1

`y-pi/4=(pi+1)(x-pi/4)`

Explanation:

For the equation we need the slope of the tangent line and the point on the curve.

Point:
`(pi/4,f(pi/4))`
`(pi/4,pi/4)`

Slope:
`f(x)=x(tanx)^2`
`f'(x)=(x)[(2)(tanx)(sec^2x)]+(1)(tan^2x)`
`f'(x)=2xtanxsec^2x+tan^2x`

`f'(pi/4)=2(pi/4)tan(pi/4)sec^2(pi/4)+tan^2(pi/4)`
`=(pi/2)(1)(sqrt2)^2+(1)^2`
`=(pi/2)(2)+1`
`=pi+1`

Point-Slope Form:
`y-y_1=m(x-x_1)`
`y-pi/4=(pi+1)(x-pi/4)`