1. ## Simplifying

I need to evaluate the derivative of the function f(t)=t/cos(t) at the point (pi/3, 2pi/3). I know that I have to find the derivative of the function and then plug pi/3 into the derivative. I did that I just don't know how to simplify my answer.

I found the derivative to be f'(x)=(sec t)(t tan t + 1)

2. Originally Posted by rawkstar
I need to evaluate the derivative of the function f(t)=t/cos(t) at the point (pi/3, 2pi/3). I know that I have to find the derivative of the function and then plug pi/3 into the derivative. I did that I just don't know how to simplify my answer.

I found the derivative to be f'(x)=(sec t)(t tan t + 1)

Oh dear: some messy-looking way to put the derivative. Won't it be easier to visualize writing $\frac{\cos t+t\sin t}{\cos^2t}=\frac{1}{\cos t}+t\tan t\,\frac{1}{\cos t}$?

Anyway, as $\cos \frac{\pi}{3}=\frac{1}{2}\,,\,\,\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ , just plug in and evaluate.

Tonio

3. Hello rawkstar
Originally Posted by rawkstar
I need to evaluate the derivative of the function f(t)=t/cos(t) at the point (pi/3, 2pi/3). I know that I have to find the derivative of the function and then plug pi/3 into the derivative. I did that I just don't know how to simplify my answer.

I found the derivative to be f'(x)=(sec t)(t tan t + 1)
$f(t) = \frac{t}{\cos t}$

Using the quotient rule:

$\Rightarrow f'(t) = \frac{\cos t + t\sin t}{\cos^2t}$, which agrees with your answer

$\Rightarrow f'(\tfrac{\pi}{3})=\frac{\frac12+\frac{\pi}{3}\fra c{\sqrt3}{2}}{\frac14}$

$=2+\frac{2\sqrt3\pi}{3}$, which won't simplify any more unless you don't mind an irrational denominator, in which case you could write is as
$2 + \frac{2\pi}{\sqrt3}$