# Thread: how do you find a dervative

1. ## how do you find a dervative

i was just wondering if anyone could help me find the derivative of a problem heres the problem:

$\displaystyle \lim_{h \to 0} \frac{\sin (\pi+h)-\sin (\pi)}{h}$

i didn't write it right i mean lim below lim there should be h arrow 0 sin (pi+h)-sin pi underline over h
how do you find the derivative of that problem and what rule do you use to solve the problem

2. Originally Posted by janemba
i was just wondering if anyone could help me find the derivative of a problem heres the problem $\displaystyle lim sin (pi+h)-sin pi[/U] h->h h$

i didn't write it right i mean lim below lim there should be h arrow 0 sin (pi+h)-sin pi underline over h
how do you find the derivative of that problem and what rule do you use to solve the problem
1. Do you know the compound angle formula for sin(A + B)?
2. Do you know the exact value of cos(pi) and sin(pi)?
3. Do you know that $\displaystyle \lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$?

3. where you put sin h =1 i mean sin (pi+h)-sin pi

4. Originally Posted by janemba
where you put sin h =1 i mean sin (pi+h)-sin pi
He means use the sum formula for $\displaystyle \sin$ on $\displaystyle \sin(\pi+h)$ then use the known values
of $\displaystyle \sin(\pi)$ and $\displaystyle \cos(\pi)$ and you will be left with something that to eveluate you will need
to know:

$\displaystyle \lim_{h\to 0}\frac{\sin(h)}{h}=1$

RonL

5. If you didn't solve yet, just like these guys told, I did like this and I think it's ok:

$\displaystyle sin(x+y) = sinxcosy + cosxsiny$
$\displaystyle \lim_{x\to 0}{\frac{sinx}{x}} = 1$

then

$\displaystyle \lim_{h\to 0}{\frac{sin(\pi+h)-sin\pi}{h}} = \lim_{h\to 0}{\frac{sin\pi cosh+cos\pi sinh - sin\pi}{h}}$
$\displaystyle = \lim_{h\to o}{\frac{sin\pi(cosh-1)+cos\pi sinh}{h}} = \lim_{h\to 0}{\frac{cos\pi sinh}{h}} = cos\pi \lim_{h\to 0}{\frac{sinh}{h}} = cos\pi = -1$

6. ok but how did you get that answer

7. Originally Posted by javax
If you didn't solve yet, just like these guys told, I did like this and I think it's ok:

$\displaystyle sin(x+y) = sinxcosy + cosxsiny$
$\displaystyle \lim_{x\to 0}{\frac{sinx}{x}} = 1$

then

$\displaystyle \lim_{h\to 0}{\frac{sin(\pi+h)-sin\pi}{h}} = \lim_{h\to 0}{\frac{sin\pi cosh+cos\pi sinh - sin\pi}{h}}$
$\displaystyle = {\color{red} \lim_{h\to o}{\frac{sin\pi(cosh-1)+cos\pi sinh}{h}}} = \lim_{h\to 0}{\frac{cos\pi sinh}{h}} = cos\pi \lim_{h\to 0}{\frac{sinh}{h}} = cos\pi = -1$
the red line is unnecessary. we know that $\displaystyle \sin \pi = 0$, so we could just eliminate the first and last terms, and go to your next line

8. Originally Posted by Jhevon
the red line is unnecessary. we know that $\displaystyle \sin \pi = 0$, so we could just eliminate the first and last terms, and go to your next line
sure Jhevon, I just wanted to be like more 'clear'...:P

Well janemba...I don't see anything else te make clear! Maybe you're not having much knowledge about trigonometry.
Visit the trigonometry section...find some of basic trigonometric identities and try to understand the Trigonometric Circle.

9. Originally Posted by janemba
ok but how did you get that answer
Perhaps if you gave simple yes or no answers to the three questions I asked in my very first post it would be easier to help you.

On the face of it, it looks like you need to go back and thoroughly revise some of the more basic material before attempting this sort of question.