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

Apr 11, 2018

Since we are talking about a tangent this means we'll want to find the gradient at that point.

#### Explanation:

$f \left(x\right) = \frac{x}{\sec \left(x\right)}$

$f \left(x\right) = x \cos \left(x\right)$

Using product rule:

$f ' \left(x\right) = \cos \left(x\right) - x \sin \left(x\right)$

Gradient at $x = \frac{\pi}{3}$ :

$f ' \left(\frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) - \frac{\pi}{3} \sin \left(\frac{\pi}{3}\right)$

$f ' \left(\frac{\pi}{3}\right) = \frac{1}{2} - \frac{\pi}{3} \times \frac{\sqrt{3}}{2}$

$f ' \left(\frac{\pi}{3}\right) = \frac{1}{2} - \frac{\pi}{2 \sqrt{3}}$

Now the equation of the tangent can be found using this gradient and the point at which the tangent is $\left(\frac{\pi}{3} , f \left(\frac{\pi}{3}\right)\right)$:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - f \left(\frac{\pi}{3}\right) = \left(\frac{1}{2} - \frac{\pi}{2 \sqrt{3}}\right) \left(x - \frac{\pi}{3}\right)$

$y = \frac{\sqrt{3} - \pi}{2 \sqrt{3}} x - \frac{\pi}{6} + {\pi}^{2} / \left(6 \sqrt{3}\right) + \frac{\pi}{3} \cos \left(\frac{\pi}{3}\right)$

$y = \frac{\sqrt{3} - \pi}{2 \sqrt{3}} x - \frac{\pi}{6} + {\pi}^{2} / \left(6 \sqrt{3}\right) + \frac{\pi}{3} \times \frac{1}{2}$

This gives you:

$y = \frac{\sqrt{3} - \pi}{2 \sqrt{3}} x + {\pi}^{2} / \left(6 \sqrt{3}\right)$