# Thread:

1. ## Derivate

Hello everyone!

I have to prove that the weak derivate of function $h(x)=sinx$ on interval $(0,2\pi)$ is function $g(x)=cosx$.
Anybody know any hint how to do this?

Thank you in advance!

2. Have a go at

$\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}$

3. Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?

4. Originally Posted by cantona
Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
Use the fact that $\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$. This will allow you to simplify the left hand side and then cancel "h".

5. Originally Posted by cantona
Ok so we get: $\displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx$. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?

If a function, h(x), is differentiable in the conventional sense, then it's weak derivative is equivalent to it's conventional derivative, g(x). That is to say any function, f(x), which which is equal to the its conventional derivative, g(x), except over some set of Lebesgue measure zero, is also a weak derivative of the function, h(x).

I'm not sure the responses reflect that this refers to Lebesgue integration.
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