Derivative - Page 2

AfterShock · #16 $Old$ September 19th, 2007, 05:40 PM

Quote:

Originally Posted by taurus $View Post$

I got another question:
Find the simplest function g(q) such that

g'(q) = sin( 5 q) +cos( 5 q)

im not sure how i should go about solving it?

First off, please place new questions in a new thread.

By finding the "integral," you're finding the "antiderivative."

$\int g'(q)\, dq = \int [\sin(5q) + \cos(5q)]\,dq = \int \sin(5q)\, dq + \int \cos(5q)\, dq = g(q)$

And thus,

$\int \sin(5q)\, dq + \int \cos(5q)\, dq = \frac{\sin(5q)}{5} - \frac{\cos(5q)}{5} + C$

We find this by, for example, letting $q = 5q$ then, $du = 5\, dq \Rightarrow \frac{1}{5} du = 5$ and thus for the first integral we have $\int \sin(5q)\, dq = \frac{1}{5}\cdot \int \sin(u)\, du = \frac{1}{5}\cdot -\cos(u) + C \Rightarrow \frac{-1}{5}\cos(5q) + C$

taurus · #17 $Old$ September 19th, 2007, 05:59 PM

hmmm is that teh only way? because it doenst quite get what i should be getting?

topsquark · #18 $Old$ September 20th, 2007, 07:40 AM

Quote:

Originally Posted by taurus $View Post$

hmmm is that teh only way? because it doenst quite get what i should be getting?

That's the only answer. You can check it by taking the derivative and seeing that it is equal to your original expression.

What answer do you think you should be getting?

-Dan