How come dy/dx $\displaystyle \sin^2(x)$ is $\displaystyle \sin(2x)$? Thanks.

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- Apr 24th 2013, 11:39 AMPazeQuick question about differentiation of sin
How come dy/dx $\displaystyle \sin^2(x)$ is $\displaystyle \sin(2x)$? Thanks.

- Apr 24th 2013, 12:59 PMMarkFLRe: Quick question about differentiation of sin
What do you get when you differentiate?

- Apr 24th 2013, 01:11 PMPazeRe: Quick question about differentiation of sin
Well I did like power rule and chain rule uhh...I'm kinda lost.. I got 2*cos(x) I believe..

- Apr 24th 2013, 01:16 PMMarkFLRe: Quick question about differentiation of sin
Yes the power and chain rules are appropriate here, but you lost sin(x) in your result...

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

What is the double-angle identity for sine? - Apr 24th 2013, 05:34 PMProve ItRe: Quick question about differentiation of sin
- Apr 24th 2013, 07:00 PMPazeRe: Quick question about differentiation of sin
Thanks Mark. I'm not too good at identifying double angle identities. I was unsure of my method due to the fact that I have a function of x squared. So I'm thinking: ''Is that the power rule or the chain rule?'' Never occured to me that when a function is squared, the function is chain rule and the 'squared' part is power rule. It was intuitive to me, but I didn't quite SEE it. Thanks for clarifying.

Prove It: although your guidance is usually of**SUPERB**quality, your description confuses me even more this time. You must be applying some identities in your calculations that I am not aware of (or I could be obtuse, sorry if I am). - Apr 24th 2013, 07:03 PMProve ItRe: Quick question about differentiation of sin
My guidance is as superb as it always is, and you are so obtuse you are almost reflex :P but I accept your apology.

Surely you know the double angle identity for cosine...

$\displaystyle \displaystyle \begin{align*} \cos{(2x)} &= \cos^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - \sin^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - 2\sin^2{(x)} \\ 2\sin^2{(x)} &= 1 - \cos{(2x)} \\ \sin^2{(x)} &= \frac{1}{2} - \frac{1}{2}\cos{(2x)} \end{align*}$

Btw no offence was intended, it's just some light-hearted ribbing :) - Apr 24th 2013, 07:32 PMPazeRe: Quick question about differentiation of sin
lol. Thanks for clarifying. I see your point now. The thing is, I just completed trigonometry in school now (final test today. It went very well :)) so it's not exactly second nature to me. These identities still have to seep in with time. I think I'm better at calculus than trigonometry, since 90% of my math knowledge is self-taught. But that, of course, means that I have a lot of gaps in my mathematical background! :(

- Apr 24th 2013, 07:41 PMMarkFLRe: Quick question about differentiation of sin
As you progress in your studies, you will get many opportunities to fortify your knowledge of pre-calculus concepts. :D