# Math Help - Sin Derivatives

1. ## Sin Derivatives

Well, this should be simpler than my previous posts =P

I need to find the derivative of this: Sin SqreRoot 2x

I use the chain rule, first to get this: Sin ((2x^1/2)) = Cos ((2x^1/2))

Then, I took the inside.

2x^1/2 = 1/2(2x)^/1/2 = X^/1/2

So, I ended up with Cos(SqreRoot2x) x 1/SqreRootX. I thought it was right, but the book is telling me the denominator is SqreRoot2x. I don't know where I went wrong.

Any help is appreciated!

2. $(\sin\sqrt{2x})'=\cos\sqrt{2x}\cdot(\sqrt{2x})'=\c os\sqrt{2x}\cdot\frac{2}{2\sqrt{2x}}=\cos\sqrt{2x} \cdot\frac{1}{\sqrt{2x}}$

3. I don't see how the derivative of SqreRoot2x is 2/2SqreRoot2x

4. $(\sqrt u)'=\frac1{2\sqrt u}\cdot u'$

It's the chain rule.