## Antiderivatives

Limits, differentiation, related rates, integration, trig integrals, etc.
GreenLantern
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Joined: Sat Mar 07, 2009 10:47 pm
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### Antiderivatives

Okay, so F(x) is the anti derivative and f(x) is the normal whatever you want to call it.
The e's and ln's have always given me problems, I just don't understand them.

$F(x) = xe^x + \pi$
$f(x) = e^x (1+x)$

$F(x) = x ln x - x$
$f(x) = ln x$

$F(x) = \sqrt{2x^2 - 1}$
$f(x) = \frac{2x}{\sqrt{2x^2 - 1}}$
My work:
$(2x^2 - 1)^{1/2}$ Take derivative
$1/2(2x^2 - 1)^{-1/2}$ Now use the chain rule
$1/2(2x^2 - 1)^{-1/2} (2x^2 - 1)$ Now I'm stuck, I just have no idea how to multiply those together; otherwise I'm pretty sure I got this one.

stapel_eliz
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The e's and ln's have always given me problems, I just don't understand them.
Do you mean that your algebra classes never covered the natural logarithm and the natural exponential?

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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### Re: Antiderivatives

I have covered both, but I can't ever remember the rules about them... I usually end up looking them up and learning them again (with varying success) every time I need them.

$F(x) = x ln x - x$
$f(x) = ln x$
My work:
$F(x) = x ln x - x$
If I'm not mistaken, I'm using the product rule because xlnx is to weird to derive from.
$F(x) = x \frac{1}{x} + ln(x) - 1$ Simplify...
$F(x) = 1 + ln(x) - 1$ Derive?
$F(x) = ln(x)$ Yes!? Point for me? It looks right...

On to round 2...
$F(x) = xe^x + \pi$
$f(x) = e^x (1+x)$
I think I'm doing the same thing? But I don't know how I'll end up with the (1+x)
$F(x) = xe^x + \pi$
$F(x) = xe^x + e^x + \pi$ That seriously looks wrong... I don't know what rule I'm using to split the xe^x and that's what I think I'm getting stuck on.

Again, the root fraction question.
$F(x) = \sqrt{2x^2 - 1}$
$f(x) = \frac{2x}{\sqrt{2x^2 - 1}}$
My work:
$(2x^2 - 1)^{\frac{1}{2}}$ Take derivative
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}}$ Now use the chain rule
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (2x^2 - 1)$ Is this yet another (longer) product rule?
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (4x) + (2x^2 - 1) \frac{1}{4}(2x^2 - 1)^{-\frac{3}{2}}$ I think that made it worse. Unless I simplify?
$2x(2x^2 - 1)^{-\frac{1}{2}} + (\frac{1}{2}x^2 - \frac{1}{4})(2x^2 - 1)^{-\frac{3}{2}}$ I realized I can simplify sooner?
Second try:
$(2x^2 - 1)^{\frac{1}{2}}$ Take derivative
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}}$ Now use the chain rule
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (2x^2 - 1)$ Simplify
$(1x^2 - \frac{1}{2})(2x^2 - 1)^{-\frac{1}{2}}$ Didn't help.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
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### Re: Antiderivatives

On to round 2...
$F(x) = xe^x + \pi$
$f(x) = e^x (1+x)$
I think I'm doing the same thing? But I don't know how I'll end up with the (1+x)
$F(x) = xe^x + \pi$
$F(x) = xe^x + e^x + \pi$ That seriously looks wrong... I don't know what rule I'm using to split the xe^x and that's what I think I'm getting stuck on.

$\frac{d}{dx}\pi=0$

factor out an $e^x$
Again, the root fraction question.
$F(x) = \sqrt{2x^2 - 1}$
$f(x) = \frac{2x}{\sqrt{2x^2 - 1}}$
My work:
$(2x^2 - 1)^{\frac{1}{2}}$ Take derivative
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}}$ Now use the chain rule
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (2x^2 - 1)$ Is this yet another (longer) product rule?
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (4x) + (2x^2 - 1) \frac{1}{4}(2x^2 - 1)^{-\frac{3}{2}}$ I think that made it worse. Unless I simplify?
$2x(2x^2 - 1)^{-\frac{1}{2}} + (\frac{1}{2}x^2 - \frac{1}{4})(2x^2 - 1)^{-\frac{3}{2}}$ I realized I can simplify sooner?
Second try:
$(2x^2 - 1)^{\frac{1}{2}}$ Take derivative
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}}$ Now use the chain rule
$\frac{1}{2}(2x^2 - 1)^{-\frac{1}{2}} (2x^2 - 1)$ Simplify
$(1x^2 - \frac{1}{2})(2x^2 - 1)^{-\frac{1}{2}}$ Didn't help.
$\frac{d}{dx}\left[\sqrt{2x^2 - 1}\right]=\frac{1}{2}\left[2x^2-1\right]^{-1/2}4x$

now simplify