Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

After using the quotient rule on

$$y=\frac{\cos x}{x}$$

I got

$$\frac{-x\sin x -\cos x}{x^2}.$$

However the answers says it should be

$$\frac{-x\sin x + \cos x}{x^2}.$$

So who's right?

share|cite|improve this question
It's always fun finding a mistake in a book. – Quang Hoang Aug 24 '14 at 19:08
So both forms are correct then? – Paul Aug 24 '14 at 19:09
@Paul Nope, you are correct, and the book answer is wrong. – m0nhawk Aug 24 '14 at 19:10
up vote 6 down vote accepted

Paul, we meet again. I see you followed my previous answer and got the right answer. However, all your book did is a little bit of algebra:

$$\frac{-x\sin x -\cos x}{x^2} =\frac{(-1)(x\sin x +\cos x)}{x^2}= -\frac{x\sin x +\cos x}{x^2}$$

I'm assuming that's what it says in the book (and you wrote it wrong in your post). If not, then the book is wrong.

share|cite|improve this answer
That's what I have just noticed, when I typed It in the mins should be like your last one. So everyone's right :) – Paul Aug 24 '14 at 19:15
Can't seem to edit that last comment, so to avoid confusion I should of said "minus" instead of "mins" – Paul Aug 24 '14 at 19:29

$y=\dfrac{\cos x}{x}$

$y'=\dfrac{\cos x'\cdot x-\cos x\cdot x'}{x^2}=\dfrac{-x\sin x\color{red}-\cos x}{x^2}$

You are right. :-)

share|cite|improve this answer

Sometimes I find it easier to rewrite in a way that allows use of the product rule.

$$y(x) = x^{-1}cos(x)$$ Now applying the power rule we have $$y'(x)=(-1)*x^{-2}cos(x)+x^{-1}(-sin(x))$$ which simplifies to $$y'(x)=-x^{-2}cos(x)-x^{-1}sin(x)$$ $$=\frac{-cos(x)}{x^{2}}-\frac{sin(x)}{x}$$ $$=\frac{-cos(x)}{x^{2}}-\frac{xsin(x)}{x^{2}}$$ $$=\frac{-cos(x)-xsin(x)}{x^{2}}$$ So as many users have already pointed out to you, you are indeed correct.

share|cite|improve this answer
\cos(x) makes $\cos(x)$, which looks better than $cos(x)$. Similarly for $\sin(x)$. – Jonas Meyer Aug 24 '14 at 19:29
Good to know. Thanks! – graydad Aug 24 '14 at 19:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.