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The question is If $$f(x)= \frac{\sec x}{x}$$, then $f'(\frac{\pi}{3}=?$ I'm kind of confused on this problem. I got $$18x-6\sqrt{3}\text{ over }\pi^2.$$

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If you are unsure of your work, it would help if you would edit to include what you did. –  Jonas Meyer Jan 24 '13 at 4:01
    
I wasn't sure what should be the numerator for your last fraction, so I just left it similar to what you already had. –  Clayton Jan 24 '13 at 4:03
    
(xsectanx-secx) over x^2 –  Captn Buzz Jan 24 '13 at 4:04
    
is that correct for the f'(x)? –  Captn Buzz Jan 24 '13 at 4:10
    
@CaptnBuzz: If you mean $$\frac{x\sec(x)\tan(x)-\sec(x)}{x^2},$$ then I believe it is correct. –  Clayton Jan 24 '13 at 4:22

1 Answer 1

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

(Use the quotient rule)

$$f'(x)=\frac{(\tan x\sec x)x-\sec x}{x^2}$$

$$f'(x)=\sec x\frac{x\tan x-1}{x^2}$$

And

$$f'(\frac{\pi}3)=\sec \frac{\pi}3\frac{\frac{\pi}3\tan \frac{\pi}3-1}{\frac{\pi^2}9}$$

$$f'(\frac{\pi}3)=2*\frac{\frac{\pi}3\sqrt3-1}{\frac{\pi^2}9}$$

So then, $$f'(\frac{\pi}3)=\frac{6(\sqrt3\pi-3)}{{\pi^2}}$$

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