Question

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

Answer 1

`y+0.5574=-3.2547(x-0.7854)`

Explanation:

First, find the point the tangent line will intersect. I'll be using decimals since there is no clean way to simplify this by any means.

`f(pi/4)=-0.5574`

The tangent line will intersect the point `(0.7854,-0.5574)`.
(Replacing `pi/4` with `0.7854`.)

To find the slope of the tangent line, find `f'(pi/4)`.

To find `f'(x)`, use quotient rule for the first term and product rule for `xtanx`.

`f'(x)=(2cosx(-sinx)e^x-e^xcos^2x)/e^(2x)-tanx-xsec^2x`

`=(-cosx(2sinx+cosx))/e^x-tanx-xsec^2x`

Plug in `pi/4`.

`f'(pi/4)=-3.2547`

Plug the point and slope into an equation in point-slope form:

`y+0.5574=-3.2547(x-0.7854)`