Question

What is the equation of the line tangent to `f(x)=xsecx` at `x=pi/3`?

Answer 1

The equation is `y - (2pi)/3 = (2 + 2/3sqrt(3)pi)(x - pi/3)`

Explanation:

We have

`f(pi/3) = (pi/3)sec(pi/3)`

`f(pi/3) = (pi/3)2`

`f(pi/3) = (2pi)/3`

Now, we find the derivative.

`f'(x) = 1(secx) + x(secxtanx)`

`f'(x) = secx + xsecxtanx`

`f'(x) = secx(1 + xtanx)`

Now find the slope of the tangent at `x = pi/3`.

`f'(pi/3) = sec(pi/3)(1 + pi/3tan(pi/3))`

`f'(pi/3) = 2(1 + pi/3(sqrt(3))/1)`

`f'(pi/3) = 2(1 + (sqrt(3)pi)/3)`

`f'(pi/3) = 2 + 2/3sqrt(3)pi`

The equation of the line is therefore:

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

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

Hopefully this helps!