derivative of sin(x) = cos(x)

[Steven G]

Hi,
I can prove that the derivative of sin(x) = cos(x) by using the definition of derivative without any problem. I can even see when I look at both graphs that this seems reasonable.
I was thinking today if there is a reason why this result would be obvious??

 

[JeffM]

What is obvious is a subjective judgment. Is this obvious? Using the series definitions of sine and cosine

[MATH]y = sin(x) \equiv \sum_{k=0}^{\infty} \dfrac{(-\ 1)^k x^{(2k+1)}}{(2k + 1)!} \implies[/MATH]
[MATH]\dfrac{dy}{dx} = \sum_{k=0}^{\infty} \dfrac{(-\ 1)^k (2k + 1)x^{\{(2k+1)-1\}}}{(2k + 1) * (2k)!} = \sum_{k=0}^{\infty} \dfrac{(-\ 1)^kx^{(2k)}}{(2k)!} \equiv cos(x).[/MATH]
If you know those definitions and what the derivatives of a sum and a power are, it is as easy as pie. Otherwise it is as mysterious as pi.

 

[Steven G]

Jeff, I agree that obvious is a subjective judgment and I need to stop saying that. What I meant by obvious (there I go again) is something a Calculus 1 student who just learned what a derivative was can use (that is from arithmetic to definition of derivative)

 

[JeffM]

This is a speculation on my part, but I wonder if you started with

[MATH]\exists \ y = f(x) \text { such that } f'(x) = \sqrt{1 - y^2} \text { for all real values of } x[/MATH]
and explored the properies of such a function, whether it would quickly be clear that f(x) = sin(x).