Question

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

Answer 1

`y-pi/2+sqrt2=(2-sqrt2)(x-pi/4)`

Explanation:

First of all, find the point the tangent line will intersect:

`f(pi/4)=(2pi)/4-sec(pi/4)=pi/2-sqrt2`

The tangent line passes through `P(pi/4,pi/2-sqrt2)`.

The part that requires calculus is finding the slope of the tangent line at `x=pi/4`, which will be equal to `f'(pi/4)`.

Differentiating the function:

`f'(x)=d/dx(2x)-d/dx(secx)`

The derivative of `2x` is `2`.

To find the derivative of `secx`, if you don't have it memorized, I'd use `secx=(cosx)^-1` then differentiate using the power and chain rule:

`d/dx(cosx)^-1=-(cosx)^-2d/dx(cosx)=(-1)/cos^2x(-sinx)`

`=1/cosx*sinx/cosx=secxtanx`

Putting this together,

`f'(x)=2-secxtanx`

And the slope of the tangent line is

`f'(pi/4)=2-sec(pi/4)tan(pi/4)=2-sqrt2(1)=2-sqrt2`.

Putting the point `P(pi/4,pi/2-sqrt2)` and slope `m=2-sqrt2` into the equation of a line:

`y-y_1=m(x-x_1)`

`y-pi/2+sqrt2=(2-sqrt2)(x-pi/4)`