find dy/dx for (sec2x)/(x^2)

Question

anonymous · Answer

anyone?

lena772 · Answer

anonymous · Answer

Ive seen that before, but ti doesn't make sense. I need it in Lamens terms.

anonymous · Answer

it*

anonymous · Answer

2sec2x(xtan2x−1)/x3

anonymous · Answer

I need to know how to solve it.

anonymous · Answer

first its rule the differentiation of
 f/g=
(f’ g - g’ f )/g2

anonymous · Answer

ya, the product rule, right?

anonymous · Answer

then compare the rule wiyh ur problem
f=sec2x
g=x^2

luigi0210 · Answer

Come on now, don't be afraid of the quotient rule.. it works better that way.

anonymous · Answer

no, I did use it, and I got the wrong answer.

luigi0210 · Answer

Did you do it right is the question?

anonymous · Answer

I well, I did substituted into the rule right, but after that it gets kinda chaotic and i think i went wrong somewhere

jdoe0001 · Answer

to get the derivative for sec(2x) you'd need to use the chain-rule I'd think

jdoe0001 · Answer

other than that, is just a plain quotient rule

luigi0210 · Answer

Here: 
$$\LARGE \frac{d}{dx}\frac {sec2x}{x^2}$$
$$\LARGE =\frac{x^2*(2sec2xtan2x)-2x(sec2x)}{(x^2)^2}$$

anonymous · Answer

Let me try that quick, and I really appreciate your guys/girls time and help so far.

luigi0210 · Answer

I know I might be wrong for 
$$\LARGE \frac{d}{dx} sec~2x$$Not sure..

anonymous · Answer

ummm, i have |dw:1393628322588:dw|