Question

How do you find the equation of the tangent line to the curve `y=secx` at `(pi/3,2)`?

Answer 1

`y = 2sqrt3(x-pi/3)+2 color(white)"XXX"`

Explanation:

The equation for the tangent line at the point `(x_1,f(x_1))` is:

`y = f'(x_1)(x-x_1)+f(x_1)`

(This is really just the point-slope form of a line in disguise!)

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The problem gives us that `x_1=pi/3` and `f(x_1)=2`. All we need to do is find `f'(x_1)` and then plug all of our values into the point-slope equation.

`f'(x)=d/dxsecx=secxtanx`

`therefore f'(pi/3)=sec(pi/3)tan(pi/3)=2*sqrt3`

Now, we need to just plug in all of these values to give us the tangent line equation:

`y = f'(x_1)(x-x_1)+f(x_1)`

`y = 2sqrt3(x-pi/3)+2`

Or if the problem prefers slope intercept form:

`y=2sqrt3x-(2pisqrt3)/3+2`