Question

How do you find the equation of the tangent line to the curve `y=2secx` at x =pi/4?

Answer 1

`y=sqrt(2)x+2sqrt(2)-sqrt(2)/2*pi`

Explanation:

The searched Tangent line has the equation

`y=mx+n`
to get we compute `f'(x)`

`f'(x)=2sec(x)tan(x)`

so we get

`f'(pi/4)=2sec(pi/4)tan(pi/4)=2sqrt(2)`
thus `m=2sqrt(2)`
To get the Point we Substitute `x=pi/4` in our function:

`f(pi/4)=2sec(pi/4)=2sqrt(2)`

so we have

`2sqrt(2)=2sqrt(2)+n`

`n=2sqrt(2)-sqrt(2)/2pi`

so the equation of our line is given by

`y=2sqrt(2)x+2sqrt(2)-sqrt(2)/2pi`