Question

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

Answer 1

See below

Explanation:

The equation of tangent line is given by

(1) `(y-y_0)=f´(x_0)(x-x_0)` where `x_0=pi/4` and `y_0=f(x_0)` and `f´(x_0)` is the value of derivative of `f` in `x_0`

Proceed:

`x_0=pi/4`
`f(pi/4)=pi^2/16cos^2(pi/4)=pi^2/32=y_0`

`f´(x)=2xcos^2x-2x^2cosxsinx=2xcos^2x-x^2sin2x`

`f´(pi/4)=cancel2·pi/4·1/cancel2-pi^2/16·1=pi/4-pi^2/16`

Apply results in (1)

`(y-pi^2/16)=(pi/4-pi^2/16)(x-pi/4)` is the tangent line equation requested