Htydjtk

First post here, so hello everyone.

Here's the problem:

Find the first derivative of: $$sqrt[3]{sin(2x)}$$

But, you can only use the difference quotient...
(i.e. the limit of $frac{f(x+h)-f(x)}{h}$ as $h rightarrow 0$)

This is a particular problem given by my prof. It's not for credit or even an assignment, she gave it as an interestingly difficult problem with some twist and turns. I've worked on it for several hours now and would like to see how others approach solving it as I'm wrapped around an axel after using double-angle ID and clearing the cubes in the numerator. Any advice is appreciated.

Thanks.

Let's use $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=sqrt[3]{sin 2(x+h)}$ and $b=sqrt[3]{sin 2x}$. Assume that $bneq 0$. This implies that $a^2+ab+b^2neq 0$ and we can divide:

$$frac{sqrt[3]{sin (2x+2h)} - sqrt[3]{sin 2x}}{h} = frac{sin (2x+2h) - sin 2x}{h} cdot frac{1}{a^2+ab+b^2}$$

Now using the identity $sin2(x+h)-sin 2x = 2sin h cos (2x+h)$, continuity of $sin$ and $cos$ and
$$lim_{~hto 0}frac{sin h}{h}=1$$

we get that the derivative is:

$$frac{2cos 2x}{3sin^{2/3} 2x}$$

I have written down all steps in detail, and I have marked the main ingredients using bold face font. This is also a nice typing exercise in Latex.

Let $pgt0$ and consider more generally the difference quotient

$$d= frac{1}{h} left(sin(2x+2h)^p - sin(2x)^p right)$$

Now for the sine of the sum we can write

$$sin(2x+2h)= sin(2x)cos(2h)+cos(2x)sin(2h) \= cos(2h)sin(2x) (1+cot(2x) tan(2h))$$

Next for $hto 0$ we have approximately

$$cos(2h)= 1- 2h^2 + O(h^4)$$
$$tan(2h) = 2h + O(h^3)$$

so that, keeping only the first order in $h$, we get for the $p$-th power

$$sin(2x+2h)^p = sin(2x)^p(1+cot(2x)2h)^p\simeq sin(2x)^p+2 h p sin(2x)^pcot(2x)\=sin(2x)^p+2 h p sin(2x)^{p-1}cos(2x)$$

where we have use the binomial formula $(1+epsilon)^p simeq 1 + p; epsilon $.

Finally we get for $hto 0$

$$d = frac{1}{h} (sin(2x)^p+2 h p sin(2x)^{p-1}cos(2x) - sin(2x)^p) \= frac{1}{h} 2 h p sin(2x)^{p-1}cos(2x)\= 2 p sin(2x)^{p-1}cos(2x)$$

For $p=frac{1}{3}$ this gives

$$d = frac{2}{3}sin(2x)^{-frac{2}{3}}cos(2x)$$

Are you allowed to use the trigonometric limit identities of
$$lim limits_{u to 0} frac{sin u}{u} = 1 hspace{.5 in} text{and} hspace{.5 in} lim limits_{u to 0} frac{1-cos u}{u} = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's proceed. The first step is to use some algebra to get rid of those pesky third roots:

begin{align*}
&lim limits_{h to 0} frac{sqrt[3]{sin(2x+2h)} - sqrt[3]{sin(2x)}}{h}\
=& lim limits_{h to 0} frac{sqrt[3]{sin(2x+2h)} - sqrt[3]{sin(2x)}}{h} cdot frac{sin(2x+2h)^{2/3} + sin(2x+2h)^{1/3}sin(2x)^{1/3} + sin(2x)^{2/3}}{sin(2x+2h)^{2/3} + sin(2x+2h)^{1/3}sin(2x)^{1/3} + sin(2x)^{2/3}}\
=&lim limits_{h to 0} frac{sin(2x+2h) - sin(2x)}{h(sin(2x+2h)^{2/3} + sin(2x+2h)^{1/3}sin(2x)^{1/3} + sin(2x)^{2/3})}
end{align*}

That denominator looks a little messy, so maybe we could call the part in parentheses
$$W(x,h) = sin(2x+2h)^{2/3} + sin(2x+2h)^{1/3}sin(2x)^{1/3} + sin(2x)^{2/3}.$$
Notice that $W(x,0) = 3sin(2x)^{2/3}$. That will come in handy later on. Now let's rewrite our limit using our $W$ shorthand and use the angle sum identity to move forward:
begin{align*}
lim limits_{h to 0} frac{sin(2x+2h) - sin(2x)}{h cdot W(x,h)} &= lim limits_{h to 0} frac{sin(2x)cos(2h) - cos(2h)sin(2x) - sin(2x)}{h cdot W(x,h)}\
&= lim limits_{h to 0} frac{cos(2h)-1}{h} cdot frac{sin(2x)}{W(x,h)} + limlimits_{h to 0} frac{-sin(2h)}{h} cdot frac{cos(2x)}{W(x,h)}\
&= lim limits_{h to 0} frac{1-cos(2h)}{2h} cdot frac{-2sin(2x)}{W(x,h)} + limlimits_{h to 0} frac{sin(2h)}{2h} cdot frac{-2cos(2x)}{W(x,h)}
end{align*}
Then, by the identities mentioned above, we have
begin{align*}
lim limits_{h to 0} frac{sqrt[3]{sin(2x+2h)} - sqrt[3]{sin(2x)}}{h} &= 0 cdot frac{-2sin(2x)}{W(x,0)} + 1 cdot frac{-2cos(2x)}{W(x,0)}\
&= frac{-2cos(2x)}{3sin(2x)^{2/3}}
end{align*}
which is the answer you're looking for.

Hint:

Use symmetry! :
$$sin(2x+2h)=sin((2x+h)+h) quad sin(2x)=sin((2x+h)-h)$$