Find y': sin x / x^2.
y' = (sin x)'*x^2-(sinx)(x^2)' / (x^2)^2
y' = cos x * x^2 - (sin x)(2x) / x^4
then what?

Question

anonymous · Answer

quotient rule here 
$$(\frac{f}{g})'=\frac{f'g-g'f}{g^2}$$

anonymous · Answer

you looking for first derivative yes?

anonymous · Answer

you have it.  factor out an x from the numerator and cancel with one of them in the denominator.  that is it

anonymous · Answer

yes. sorry - i should have said in the beginning i was using the quotient rule

anonymous · Answer

get 
$$\frac{x \cos(x)-2\sin(x)}{x^3}$$

anonymous · Answer

ok. this is where I am stuck: how do you get x cos x? To the point, how does the x move in front of cos x? wouldn't it be just cos x once the x  is factored out?

anonymous · Answer

your numerator was 
$$\cos(x)x^2-2x\sin(x)=x^2\cos(x)-2x\sin(x)=x(x\cos(x)-2\sin(x))$$

anonymous · Answer

just the commutative law and factoring.  this allows you to cancel one of the "x"s from the denominator

anonymous · Answer

now i see my error: (sin x)' does not equal cos. it equals cos x.