How do you find the equation for the tangent line to sec2x at x=π3?

Question

ti2n1i2mt0l7 · Accepted Answer

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 $y = \sec^{2} x$ and we are seeking tangent at point $x = \frac{π}{3}$ i.e. at
$(\frac{π}{3}, \sec^{2} (\frac{π}{3}))$ or $(\frac{π}{3}, 4)$
As $f (x) = \sec^{2} x, f^{'} (x) = 2 \sec x \times \sec x \tan x = 2 \sec^{2} x \tan x$
and slope of tangent is $f^{'} (\frac{π}{3}) = 2 \sec^{2} (\frac{π}{3}) \tan (\frac{π}{3}) = 8 \sqrt{3}$
and hence equation of tangent is the equation of line passing through $(\frac{π}{3}, 4)$ and slope $8 \sqrt{3}$
i.e. $y - 4 = 8 \sqrt{3} (x - \frac{π}{3})$
or $8 \sqrt{3} x - y + 4 - \frac{8 π}{3} = 0$

How do you find the equation for the

Answered question

Answer & Explanation

New Questions in Calculus 1