Derivative of secx where x = pi/3

[tmt1]

The derivative of secx is

$$\d{y}{x} secx =secx tanx $$

But if $$x = \frac{\pi}{3}$$, then $$secx = 2 $$ and the derivative of a constant is 0.

And $$sec\frac{\pi}{3} tan\frac{\pi}{3}$$ is equal to $$\frac{3}{2}$$

So what is the derivative of $$secx$$ where $$x = \frac{\pi}{3}$$?

 

[SuperSonic4]

The derivative of secx is

$$\d{y}{x} secx =secx tanx $$

But if $$x = \frac{\pi}{3}$$, then $$secx = 2 $$ and the derivative of a constant is 0.

And $$sec\frac{\pi}{3} tan\frac{\pi}{3}$$ is equal to $$\frac{3}{2}$$

So what is the derivative of $$secx$$ where $$x = \frac{\pi}{3}$$?

The derivative is the rate of change at a given point.

When you did $$\dfrac{d}{dx}\sec(x)$$ you changed from your line itself to the rate of change of the line (i.e. how fast the slope is changing). When you do [math]\sec\left(\dfrac{\pi}{3}\right) \tan \left(\dfrac{\pi}{3}\right) [/math] you're finding out how fast the gradient is changing which happens to be $$2\sqrt{3}$$

When you say [math]f(x) = \sec \left(\dfrac{\pi}{3}\right)[/math] you're describing a horizontal line which has no gradient (and is so 0)

If you don't understand I'll try and be clearer

edit: here's a graph (using the MHB widgets (Talking))
plot sec'('x')',tan'('x')'sec'('x')',sec'('pi'/'3')' between x'='-2pi and x '=' 2pi - Wolfram|Alpha

Purple is the derivative and and blue is the original function - note how purple (the derivative) can be zero. The line y = 2 is sec(pi/3)