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The question asks to find the derivative of the function $1-\cos(x)\sin(x)$, and I thought maybe using some derivative rules I could, but I don't know where to start.

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Apply the rules: how to differentiate $f-g$, constant, then $f\cdot g$, finally you also need $\sin'=\cos$ and $\cos'=-\sin$. –  Berci Oct 15 '12 at 0:56
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3 Answers 3

up vote 4 down vote accepted

Are you familiar with the product rule? $$ f(x)=u(x)v(x)\\ f'(x)=u'(x)v(x) + u(x)v'(x) $$ In your case $u(x)=\cos(x), \ v(x)=\sin(x)$.

EDIT Sorry again for the typos, Here's what you should do. The derivative of constant is always 0, derivative of $-h(x)$ is $-h'(x)$, the derivative of a product of functions is above, the derivative of $\cos'(x)=- \sin(x), \ \sin'(x)=\cos(x)$

Can you handle it now?

EDIT 2: Sorry for doing it again, there is a different way of solving the problem if you notice that

$$ -\cos(x) \sin(x) = -\frac{2}{2}\cos(x) \sin(x)=-\frac{1}{2}\sin(2x) $$ and then use the chain rule: $\frac{du(v(x))}{dx} = u'(v(x))v'(x)$ and then use the derivative of the sin function

EDIT 3: OK here is the solution: $$ f'(x)=(-\cos(x)\sin(x))'_{x}=(-\frac{1}{2}\sin(2x))'_{x}=-\frac{1}{2}(2x)'_{x}\cos(2x)=-\cos(2x) $$

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you mean the product rule –  Jean-Sébastien Oct 15 '12 at 0:53
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@Alex : that is the product rule. We don't want to confuse the OP. Also, you can use "\sin" and "\cos". –  Stefan Smith Oct 15 '12 at 0:54
    
Well, yes, you can rewrite it using $\sin(2x) = 2\sin(x)\cos(x)$ –  Berci Oct 15 '12 at 0:57
    
Sorry for the typo, will update it now –  Alex Oct 15 '12 at 0:58
    
I am not sure how you got the 2 in there? –  Jillian Johnson Oct 15 '12 at 1:00
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Using the well known trigonometric identity,

$$\sin(2x)=2\sin(x)\cos(x)$$

We can say that,

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

and the derivative would be:

$$y'=0-\frac{2\cos(2x)}{2}=-\cos(2x)$$

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Upvoted for greatness. However, it won't help him in the long run because if he's asking this he clearly doesn't know about the product rule. –  Damieh Oct 15 '12 at 15:10
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$$(1 - \sin(x)\cos(x))' = \left(1 - \frac12 \sin(2x)\right)' = -\cos(2x).$$ Implicitly, I used the chain rule here, letting $f(x) = \sin(x)$ and $ g(x) = 2x$.

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Where did you get the -(1/2)sin(2x)?? –  Jillian Johnson Oct 15 '12 at 1:13
    
Sorry, I dont know the chain rule. But thank you for helping. –  Jillian Johnson Oct 15 '12 at 1:16
    
By using $\sin(2x) = 2\sin(x)\cos(x).$ –  Ian Mateus Oct 15 '12 at 1:29
    
Got it, thanks! –  Jillian Johnson Oct 15 '12 at 1:49
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