1. Find the derivative of...

Find the derivative for $\displaystyle f(x)=\frac{cos(2x)}{x}.$

I'm actually supposed to find the double derivative but I can't find the first derivative first.

Anyone care to reveal the steps to find out the derivative of the above.

I've tried both using the quotient and product rule but have miserably failed in trying to do so.

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:

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

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

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.

3. You can either use the quotient rule with $\displaystyle u = \cos(2x)$ (and don't forget the chain rule) and $\displaystyle v = x$

Or you may use the product rule if you say that $\displaystyle f(x) = x^{-1} \cdot \cos(2x)$

Both methods are exactly the same, it just comes down to whichever rule you prefer

4. 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:

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

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

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.
Thank you for all your help so far. I appreciate it and I follow what you've done but how do I find the derivative of the derivative (your answer). I also keep getting that one wrong.
i.e. I also need to find the second derivative but don't know how to.

5. The chain rule allows you to differentiate the function from the inside out which is absolutely necessary when finding any derivative. Once you've differentiated according to the chain rule, you've found the first derivative. You find the second derivative by differentiating the first derivative (using the chain rule if necessary).

I.e. the derivative of the derivative is the second derivative.

$\displaystyle f''(x) = [f'(x)]'$

The same applies for any derivative. Any derivative can be obtained by differentiating its primitive function:

$\displaystyle f^{(n)} (x) = [f^{(n-1)}(x)]'$

Here you have to differentiate $\displaystyle \frac{-2x\sin {2x} - \cos{2x}}{x^2}$ to find the second derivative, $\displaystyle f''(x)$ Now that's complicated =)

6. Originally Posted by mrfour44
Here you have to differentiate $\displaystyle \frac{-2x\sin {2x} - \cos{2x}}{x^2}$ to find the second derivative, $\displaystyle f''(x)$ Now that's complicated =)
Could you show me the step-by-step working out for it please?

I really want to see what I was supposed to do and where I went wrong.

7. Sure, no problem.

$\displaystyle f(x)= \frac{\cos{2x}}{x}}$

We are to find the first derivative in x, $\displaystyle f'(x)$, and the second derivative in x, $\displaystyle f''(x)$.

The first derivative is obtained by differentiating the function using the quotient rule, all the while taking care that the cosine is differentiated using the chain rule as well:

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

The cosine is differentiated using the chain rule, therefore $\displaystyle (\cos{2x})' = - 2\sin{2x}$, and we include that.

The first derivative is therefore:
$\displaystyle f'(x) = \frac{-2x\sin{2x} - \cos{2x}}{x^2}$

The second derivative is obtained by differentiating the first derivative in x:
$\displaystyle f''(x) = (f'(x))' = \left(\frac{-2x\sin{2x} - \cos{2x}}{x^2}\right)'$

Analogously:
$\displaystyle f''(x) = \frac{(-2x\sin{2x} - \cos{2x})'\cdot x^2 - (x^2)'\cdot(-2x\sin{2x} - \cos{2x})}{x^4}$
$\displaystyle = \frac{x^2(-(2x\sin{2x})'-(\cos{2x})') - 2x(-2x\sin{2x}-\cos{2x})}{x^4} =$
$\displaystyle = \frac{x^2(-2\sin{2x}-4x\cos{2x}+2\sin{2x})+2x(2x\sin{2x}+\cos{2x})}{x^4 }=$,

$\displaystyle f''(x) = \frac{-4x^2\cos{2x} + 4x\sin{2x} + 2\cos{2x}}{x^3}$

After reducing the fraction by an 'x'.

8. Originally Posted by Joker37
Find the derivative for $\displaystyle f(x)=\frac{cos(2x)}{x}.$

I'm actually supposed to find the double derivative but I can't find the first derivative first.

Anyone care to reveal the steps to find out the derivative of the above.

I've tried both using the quotient and product rule but have miserably failed in trying to do so.
Another way to apply the Chain Rule is

$\displaystyle u=2x \Rightarrow \frac{d}{dx}\frac{cos2x}{x}=\frac{d}{dx}\frac{cosu }{\frac{1}{2}u}=2\frac{d}{dx}\frac{cosu}{u}=2\frac {du}{dx}\frac{d}{du}\frac{cosu}{u}=4\frac{d}{du} \frac{cosu}{u}$

$\displaystyle =4\frac{u\frac{d}{du}cosu-cosu\frac{d}{du}u}{u^2}=4\frac{u(-sinu)-cosu}{u^2}$

$\displaystyle =-\frac{2xsin2x+cos2x}{x^2}$

Then to get the 2nd derivative

$\displaystyle -4\frac{d}{dx}\frac{usinu+cosu}{u^2}=-4\frac{du}{dx}\frac{d}{du}\frac{usinu+cosu}{u^2}=-8\frac{d}{du}\frac{usinu+cosu}{u^2}$

Now apply the Quotient Rule on the single fraction,
or independently for 2 fractions.