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before I ask for anything I must admit I'm working hard to understand this beautiful subject. Thanks in advance. I want to get a derivative of: $\cos(x)\sin(x)$ The solution is the following:

$$ \frac d{dx}(\cos(x) \sin(x)) = \cos(2 x)$$

Where should I start?

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Do you know the product rule? $\frac{\mathrm d}{\mathrm dx} (f(x) g(x)) = f(x) \frac{\mathrm d}{\mathrm dx} g(x) + g(x) \frac{\mathrm d}{\mathrm dx} f(x)$. – Rahul Nov 10 '12 at 3:36
Of course, but I did not get a cos(2x) Thanks. – Rakisbro Nov 10 '12 at 3:50
up vote 3 down vote accepted

One way is to recall the identity $\sin 2x=2\sin x\cos x$, which can be rewritten as $$\sin x\cos x=\frac{1}{2}\sin 2x.$$ Now differentiate, not forgetting to use the Chain Rule.

An alternative is to differentiate $\cos x\sin x$ immediately, using the Product Rule. We get $(\cos x)(\cos x)+(-\sin x)(\sin x)=\cos^2 x-\sin^2 x$. Then recall the identity $\cos 2x=\cos^2 x-\sin^2 x$.

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I did not know $cos(2x) = cosx^2 - senx^2$ identity. Thanks – Rakisbro Nov 10 '12 at 3:58
Maybe you know $\cos 2x=2\cos^2 x-1$. The right-hand side is $\cos^2 x+\cos^2 x-1$. Replace one of the $\cos^2 x$ by $1-\sin^2 x$. – André Nicolas Nov 10 '12 at 4:04

Do you know the rules for taking the derivatives of products?

$$\frac d{dx}(uv) = u'v + uv'$$

Set $u = \cos(x)$ and $v = \sin(x)$ and find their derivatives using this rule.

Then you will need to use a Double-Angle Formula to get your desired result.

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