Understanding the Quotient Rule for Derivatives of 3/x^2

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Homework Statement


3/x^2

when I take the quotient rule ,
I get:
(0*x^2-3*2x)/(x^2)^2
isn't that -6x/x^4 or -6/x^3
My calculator says -6 and so it is, but why and what am I missing?
 
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ok From what i see you are right.the answer is -6x^-3.

but at what point are you evalusting.did they give you a point or did you just type on your calculator.you have to have value of x.to type on your calculator
 
I originally had to differentiate f(t)=cos^-1(3/t^2)
my answer was f'(t)=-1/sqrt(1-(3/t^2)^2)*-6t
f'(t)=6t/sqrt(1-9/t^4)
I got the "-6t" part wrong, it was supposed to be just -6.

when I differentiate just 3/x^2 into my calculator and it comes out -6
on my ti89 calc. I go to math, calculus, differentiate and type in:
d(3/x^2,x) and get -6
 
Instead of using the quotient rule, you might find it easier to evaluate
3 x^-2
 
Oky, thanks. I see my problem.
I should have looked into the mirror sooner, the problem was there all along!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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