given a derivative find original function

**ives** · November 19th 2009, 01:06 PM

given a function f(x) , find a g(x) such that g'(x)=f(x)

(1) 1/(4+x^2)

(2) cos x * sin x

great appreciate for any help

**Drexel28** · November 19th 2009, 01:09 PM

Originally Posted by ives

given a function f(x) , find a g(x) such that g'(x)=f(x)

(1) 1/(4+x^2)

(2) cos x * sin x

great appreciate for any help

A) I believe that there is a name for this...hmm...integration?

B) I don't think this is analysis, topology, or differential geometry. Try the calculus section next time.

1. $x=2\tan\vartheta$

2. Note that $\bigg[\sin(x)\bigg]'=\cos(x)$

**ives** · November 19th 2009, 01:41 PM

i kno it's integration,, but i haven't learned that yet,,,,,..

**redsoxfan325** · November 19th 2009, 05:55 PM

If you can't integrate, then it's basically guess and check, though you should at least know what general types of functions you should be looking for.

**mr fantastic** · November 20th 2009, 01:42 AM

Originally Posted by ives

i kno it's integration,, but i haven't learned that yet,,,,,..

Then you should learn it and attempt simpler questions before attempting problems like these ones. Because if what you say is true then the solutions to these questions will be meaningless.

**Haven** · November 20th 2009, 01:54 AM

They do anti-derivatives in Calculus 1, in universities. They usually don't learn about Riemann sums and such until the next course in Calculus. These question were probably assigned or done in lecture and he doesn't understand them.

If your given a function $f(x)$ we're basically trying to find a function $F(x)$ such that $F'(x) = f(x)$

Knowing this we can reverse some of the easier differential rules.

Like the power rule. if $f(x) = x^n$ then $F(x) = \frac{1}{n+1}x^{n+1} + C$
Since the derivative of a constant is 0.

Or with cos and sin. if $f(x) = \cos{x}$ then $F(x) = \sin{x} + C$
if $f(x) = \sin{x}$ then $F(x) = -\cos{x} + C$

For the first question note that: $\frac{d}{dx}(\arctan{\frac{x}{a}}) = \frac{1}{1 + \frac{x^2}{a^2}}$
For the second question, a helpful identity is that $\sin{2x} = 2\sin{x}\cos{x}$

**Scott H** · November 20th 2009, 03:00 AM

Originally Posted by Haven

For the first question note that: $\frac{d}{dx}(\arctan{\frac{x}{a}}) = \frac{1}{1 + \frac{x^2}{a^2}}$

Correction: $\frac{d}{dx}\arctan{\frac{x}{a}}=\frac{1}{1+\frac{ x^2}{a^2}}\cdot\frac{1}{a}.$

P.S. I like your sig!

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