Question

For `f(x)=xcos^2x` what is the equation of the tangent line at `x=pi/4`?

Answer 1

`y = (2-pi)/4 x + pi^2/16`

Explanation:

For a line, we can calculate "point-slope" form. We can calculate the point:
`f(pi/4) = pi/4 * cos^2(pi/4) = pi/4 * (sqrt(2)/2)^2 = pi/8`
So it intercepts `(pi/4, pi/8)`.

The slope we calculate through the derivative:
`f'(x) = cos^2(x)-2xcos(x)sin(x)`
`f'(pi/4) = 1/2 - pi/2 * 1/2 = (2-pi)/4`

Therefore, we can use point-slope form:
`y - y_1 = m(x-x_1)`
`y-pi/8 = (2-pi)/4 * (x-pi/4)`

`y = (2-pi)/4 x + pi/8 - (2pi - pi^2)/16`
`y = (2-pi)/4 x + pi^2/16`

`y approx -0.285 x + 0.617`