Originally Posted by

**mrfour44** Judging by your function, you are to find the derivative in $\displaystyle x$, right?

So, you have to differentiate using *the chain rule* and the quotient rule. The numerator is a composition of the sine function and of the linear function (a function of a function). See:

(For lack of confusion, I will use 't' instead of 'x' as the dummy variable.)

$\displaystyle f_1(t) = \cos{t} $ and $\displaystyle f_2 (t) = 2t $

Once those two are joined as a composition, then we get:

$\displaystyle (f_1 \circ f_2)(t) = \cos{(2t)} $, or in your case $\displaystyle \cos{(2x)} $.

This function is then divided by another, and that is why you need the quotient rule.

The chain rule shows us that:

$\displaystyle \frac{d(f(x))}{dx} = \frac{d(f(x))}{dz} \cdot \frac{dz}{dx} $

Where, in your case, $\displaystyle z = 2x $.

The derivative of $\displaystyle \cos{2x}$ in $\displaystyle x$ is:

$\displaystyle \frac{d}{dx}\cos{2x} = -2\sin{2x} $

You first differentiate the cosine $\displaystyle (\cos{2x})' = (- \sin{2x})\cdot\dots $ and then the function inside and multiply the original, $\displaystyle (\cos{2x})' = (-\sin{2x})\cdot(2x)'=-2\sin{2x}$.

Then, by using the quotient rule:

$\displaystyle \left(\frac{f_1 (x)}{f_2 (x)}\right)' = \frac{f'_1 (x)f_2(x) - f_1(x)f'_2(x)}{f_2^2(x)}$

In your case:

$\displaystyle \left(\frac{\cos{2x}}{x}}\right)' = \frac{-2(\sin{2x})\cdot x - 1\cdot\cos{2x}}{x^2} = \frac{ - 2x\sin{2x} - \cos{2x}}{x^2} $

Maybe the expression can be simplified, but it looks nice either way =)

Good luck and remember the chain rule when you differentiate a composition of functions.