# Math Help - ...Anti-Derivatives?

1. ## ...Anti-Derivatives?

I guess this are called Anti-Derivatives, which is strange since we never discussed it in class.

Anyways, I'm given a function's derivative and they want me to find the actual function. It's not too bad, but then they ask us to find the function through a certain point, and that's when I get lost

Here's the questions I'm stuck on...

Find the function with the given derivative whose graph passes through the given point.

A) 1/xSquared + 2x, Point = (-1,1)
B) 8-CscSquared, Point = (Pi/4,0)
C) SecXTanX - 1, Point (0,0)

Any help is appreciated!

2. Originally Posted by Super Mallow
I guess this are called Anti-Derivatives, which is strange since we never discussed it in class.

Anyways, I'm given a function's derivative and they want me to find the actual function. It's not too bad, but then they ask us to find the function through a certain point, and that's when I get lost

Here's the questions I'm stuck on...

Find the function with the given derivative whose graph passes through the given point.

A) 1/xSquared + 2x, Point = (-1,1)
B) 8-CscSquared, Point = (Pi/4,0)
C) SecXTanX - 1, Point (0,0)

Any help is appreciated!
how far did you get with anti-derivatives? they weren't discussed at all? nothing?

if so, i guess you have no idea what the power rule is. here's what you do. look up all your derivative formulas. the things you have here will be the right side of the equation. the anti-derivative will be on the left. for example,

you should find that: $\frac d{dx}\sec x = \sec x \tan x$

so the anti-derivative of $\sec x \tan x$ is $\sec x + C$ since sec(x) is what you derived to get the sec(x)tan(x) so taking the antiderivative of that brings you back to sec(x). the + C is a constant. remember that when we take the derivative, if there is a constant, it's derivative is zero, so we put + C just in case there was a constant that got wiped out by the differentiation.

continuing with this example:

we have $\sec x \tan x - 1$

the anti-derivative of this is: $\sec x - x + C$ (when we differentiate x we get 1).

we want this to pass through (0,0), which means we want when x = 0, y = 0

so, plug that in: $\sec 0 - 0 + C = 0$

now solve for C

we get: $1 + C = 0 \implies C = -1$

so the antiderivative is: $\sec x - x - 1$

try the others

3. Yeah, we had no talks about Anti-Derivatives whatsoever. My roomate told me that they are preparing us for it in the upcoming weeks.

This may be a dumb question and may be why I don't understand it...what is C? Why did we put it in there?

4. Originally Posted by Super Mallow
Yeah, we had no talks about Anti-Derivatives whatsoever. My roomate told me that they are preparing us for it in the upcoming weeks.

This may be a dumb question and may be why I don't understand it...what is C? Why did we put it in there?
did you read my post, or did you not understand it?

Originally Posted by Jhevon
...the + C is a constant. remember that when we take the derivative, if there is a constant, it's derivative is zero [it gets wiped out], so we put + C just in case there was a constant that got wiped out by the differentiation....

5. My bad: I didn't really understand it

6. Originally Posted by Super Mallow
My bad: I didn't really understand it
ok, let's add a little example shall we.

let's say we had a function, a simple one, like y = x + 1

now, take it's derivative, we get:

y' = 1

now, if we take the anti-derivative, we should get back the original function.

but let's say we said the anti-derivative was x. that would not be true, there was a 1 there, so we would have something missing from our original function. now if we said, the anti-derivative is x + C, that is fine, there is nothing missing, but the specific value of C, which we can find if we are given the appropriate clues. so that's why when finding anti-derivatives, we add an arbitrary constant, just in case there was a constant there that got wiped out by differentiation