Question

How do you find the equation for the tangent line to `sec^2x` at `x=pi/3`?

Answer 1

Equation of tangent is `8sqrt3x-y+4-(8pi)/3=0`

Explanation:

The slope of tangent to a curve `y=f(x)` at a point `x=x_0` is given by `f'(x_0)`.

Here the function is `y=sec^2x` and we are seeking tangent at point `x=pi/3` i.e. at `(pi/3,sec^2(pi/3))` or `(pi/3,4)`.

As `f(x)=sec^2x`, `f'(x)=2secx xx secxtanx=2sec^2xtanx`

and slope of tangent is `f'(pi/3)=2sec^2(pi/3)tan(pi/3)=8sqrt3`

and hence equation of tangent is the equation of line passing through `(pi/3,4)` and slope `8sqrt3`

i.e. `y-4=8sqrt3(x-pi/3)`

or `8sqrt3x-y+4-(8pi)/3=0`