# 1st derivatives of $f(\alpha) = \frac{\sin(2\alpha)}{\sin(\alpha+1)}$

Could someone help me out with the following? I have to get a maximum using the derivative

$$f(\alpha) = \frac{\sin(2\alpha)}{\sin(\alpha+1)}$$

$$f(\alpha) = \sin(2\alpha) \cdot (\sin(\alpha+1))^{-1}$$

$$f'(\alpha) = \sin(2\alpha) \cdot ((\sin(\alpha+1))^{-1})' + (\sin(2\alpha))' \cdot (\sin(\alpha+1))^{-1}$$

$$f'(\alpha) = \sin(2\alpha) \cdot \color{red}{\cdots} + 2 \cos(2\alpha) \cdot (\sin(\alpha+1))^{-1}$$

I can't get any furher then this

• Wait, are your $\alpha$'s supposed to be $x$'s? Otherwise, what are the $\alpha's$?
– LASV
Commented Dec 12, 2013 at 18:57
• @LASV yes, edited ;) Commented Dec 12, 2013 at 18:57
• @Mazzy, edit also the title Commented Dec 12, 2013 at 18:58
• What's the derivative of $\;\sin 2\alpha\;$ ? What the derivative of $\;\sin(\alpha+1)\;$ ? Now apply directly the quotient rule... Commented Dec 12, 2013 at 18:59
• @MichaelHardy I went with the OP's notation. And interpreting $\sin^{-1} x$ as $\frac{1}{\sin x}$ is perfectly legitimate, and not so unusual. After all, when reading $\sin^2 x$ for $(\sin x)^2$, nobody twitches an eye and says that must mean $\sin (\sin x)$ and nothing else, do they? Commented Dec 12, 2013 at 19:07

Use the quotient rule: $$\left(\frac{f(x)}{g(x)} \right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$ here we have $f(x)=\sin(2x)$ and $g(x)=\sin(x+1)$. So we have $$\frac{2\cos(2x)\sin(x+1)-\sin(2x)\cos(x+1)}{\sin^2(x+1)}$$ If you are looking for a maximum, set this equal to zero and solve.

EDIT. As it has been suggested, the intended problem was $$\left(\frac{\sin(2x)}{\sin(x)+1} \right)'$$ (always a good idea to include parenthesis around function arguments where confusion could arise!) So that the derivative of the denominator is $\left(\sin(x)+1\right)'=\cos(x)+0=\cos x$. That would give the solution: $$\frac{2\cos(2x)\left(\sin(x)+1\right)-\sin(2x)\cos(x)}{\left(\sin(x)+1\right)^2}$$ which matches their solution.

• Just to add a confirmation that is not needed: the answer is correct. Conceivably they expanded the $\sin(2x)$ as $2\sin x\cos x$, in which case we could get a $\cos x$ at the end of the top part, but we would also have a $2$, and a $\sin x$. Commented Dec 12, 2013 at 19:15
• @Mazzy Their solution is incorrect. Moreover, they have also written $(\sin \alpha+1)$ when they meant to write $\sin(\alpha+1)$, a double error! (Triple if you include the same problem with how it is written in the denominator) Commented Dec 12, 2013 at 19:20
• @Mazzy: Don't call! The solution in the link is correct. So is the solution above. They are solutions to different problems! You stated the problem wrongly. The denominator is supposed to be $\sin\alpha+1$, not $\sin(\alpha+1)$. Good thing too, since with your version solving derivative equal to $0$ could only be done numerically. Commented Dec 12, 2013 at 19:30
• You are welcome. Pity, maybe, but things like $\cos x$, $\tan t$ are standard. What they really should have done is write $1+\sin\alpha$. Commented Dec 12, 2013 at 19:44
• My observation comes from too many years making up questions, both for courses and contests. After one decides on the basic problem, the hard work is examining the thing symbol by symbol, looking for places where misinterpretation is conceivable. After all, who wants to be sued? Commented Dec 12, 2013 at 19:49

As mathematics2x2life has pointed out, the quotient rule can be used. However, I will address directly the question of what to do with $$\frac{d}{d\alpha} (\sin(\alpha+1))^{-1}.$$

You can say it's $\dfrac{d}{d\alpha} u^{-1}$, so that it's $$\frac{d}{d\alpha} u^{-1} = (-1)u^{-2}\cdot\frac{du}{d\alpha} = -u^{-2}\cdot\frac{d}{d\alpha} \sin(\alpha+1) = \frac{-1}{u^2}\cdot\cos(\alpha+1).$$

After that you would multiply by $\dfrac{d}{d\alpha}(\alpha+1)$, but you'd just be multiplying by $1$, so that doesn't change anything. Then you have $$\frac{-1}{(\sin(\alpha+1))^2}\cdot\cos(\alpha+1).$$