How do you differentiate f(x)=(tanx-1)/secx at x=pi/3?

Dec 17, 2016

$\frac{1 + \sqrt{3}}{2}$

Explanation:

Rewrite in sine and cosine.

$f \left(x\right) = \frac{\sin \frac{x}{\cos} x - 1}{\frac{1}{\cos} x}$

$f \left(x\right) = \frac{\frac{\sin x - \cos x}{\cos} x}{\frac{1}{\cos} x}$

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

We differentiate this using $\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$ and $\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$.

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

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

We now evaluate $f ' \left(\frac{\pi}{3}\right)$:

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

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

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

Hopefully this helps!