3

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?

Michael Albanese
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Abby
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3 Answers3

6

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.

Shahar
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  • That's what I have just noticed, when I typed It in the mins should be like your last one. So everyone's right :) – Abby 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" – Abby Aug 24 '14 at 19:29
2

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

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

You are right. :-)

rae306
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1

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.

graydad
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