need help understanding d/dx(x^(1/2))

Hi,
I'm having problems understanding the derivation of fraction exponents like this for example:
d/dx(x^(1/2))

Obviously I have more complicated problems than this to evaluate, but a basic explanation would be helpful.
The book I'm using for some reason skips talking about this....

The example I'm looking at on my homework is this:
d/dx((4/z^(1/3))-(z^(1/4)))

4 over the 3rd root of z, minus z to the 4th root.

Thanks much,
sig

Quote:

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Originally Posted by sig

Hi,
I'm having problems understanding the derivation of fraction exponents like this for example:
d/dx(x^(1/2))

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There is one basic power rule: $\displaystyle \frac{d}{dx}\left(x^{\alpha}\right)=\alpha x^{\alpha-1} $

Thus $\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{ 2} x^{-\frac{1}{2}} $

Quote:

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Originally Posted by Plato

There is one basic power rule: $\displaystyle \frac{d}{dx}\left(x^{\alpha}\right)=\alpha x^{\alpha-1} $

Thus $\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{ 2} x^{-\frac{1}{2}} $

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I must be making mistakes when I evaluate then, because I know the power rule.
For instance my answer sheet says

$\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=-\frac{1}{2x^{\frac{1}{2}}}$

I would think the x^(-1/2) would go on top of the fraction, and I don't know how the negative is pulled out of the exponent into the front of the answer. (Nerd)

Quote:

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Originally Posted by sig

my answer sheet says

$\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=-\frac{1}{2x^{\frac{1}{2}}}$

I would think the x^(-1/2) would go on top of the fraction, and I don't know how the negative is pulled out of the exponent into the front of the answer. (Nerd)

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Your answer sheet is wrong,

It should be $\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{ 2x^{\frac{1}{2}}}$

Recall that $\displaystyle \left(x^{-\frac{1}{2}}\right)=\frac{1}{x^{\frac{1}{2}}}$

Quote:

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Originally Posted by Plato

Your answer sheet is wrong,

It should be $\displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{ 2x^{\frac{1}{2}}}$

Recall that $\displaystyle \left(x^{-\frac{1}{2}}\right)=\frac{1}{2x^{\frac{1}{2}}}$

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Oh I'm sorry, the negative was a part of a different problem.
Oh, and now I see how the numbers end up on bottom; the negative exponent.

OK thank you very much, I'll try some equations again and see if I can understand better.

-sig

Hello, I'm back.
After looking and working on this equation,

$\displaystyle \frac{d}{dz}\left((4/z^{\frac{1}{3}}\right)-(z^\frac{1}{4}))$

I've come to the pre-simplified answer of:

-(4/3z^(4/3))-(1/(4z^(3/4))

And when I look the answer up on WRA, it gives the simplified answer of:

-((3z^(7/12)+16)/(12z^(4/3))

and I'm not seeing how this simplification is working.

Can anyone help explain?

Thanks..
-sig

Quote:

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Originally Posted by sig

Hello, I'm back.
After looking and working on this equation,

$\displaystyle \frac{d}{dz}\left(4/z^{\frac{1}{3}}\right)$

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That is $\displaystyle \frac{d}{dz}\left(4z^{-\frac{1}{3}}\right)=\frac{-4}{3}z^{\frac{- 4}{3}}$

Quote:

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Originally Posted by Plato

That is $\displaystyle \frac{d}{dz}\left(4z^{-\frac{1}{3}}\right)=\frac{-4}{3}z^{\frac{- 4}{3}}$

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Yes I got that much.
which equals

-(4/3z^(4/3))

But I don't see how this:

-(4/3z^(4/3)) - (1/4z^(/34))

turns into this:

- ((3z^(7/12)+16)/(12z^(4/3)))


Sorry for not entering this into special text- is there a way to convert without manually coding it?

Quote:

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Originally Posted by sig

Hello, I'm back.
After looking and working on this equation,

$\displaystyle \frac{d}{dz}\left((4/z^{\frac{1}{3}}\right)-(z^\frac{1}{4}))$

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Those parentheses are a little confusing. Do you mean $\displaystyle \frac d{dz}\left[\frac4{z^{1/3}}-z^{1/4}\right]$?

We have

$\displaystyle \frac d{dz}\left[\frac4{z^{1/3}}-z^{1/4}\right]$

$\displaystyle =\frac d{dz}\left[4z^{-1/3}}-z^{1/4}\right]$

$\displaystyle =-\frac43z^{-4/3}-\frac14z^{-3/4}$

$\displaystyle =-\frac4{3z^{4/3}}-\frac1{4z^{3/4}}$

$\displaystyle =-\frac4{3z^{16/12}}-\frac1{4z^{9/12}}$

and, combining fractions,

$\displaystyle =-\frac{4(4)}{\left(3z^{16/12}\right)(4)}-\frac{3z^{7/12}}{\left(4z^{9/12}\right)\left(3z^{7/12}\right)}$

$\displaystyle =\frac{-16-3z^{7/12}}{12z^{4/3}}$

$\displaystyle =-\frac{3z^{7/12}+16}{12z^{4/3}}$

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$\displaystyle =-\frac4{3z^{16/12}}-\frac1{4z^{9/12}}$

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Yes that's right and this above is where I can get to. But here below I don't understand

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and, combining fractions,

$\displaystyle =-\frac{4(4)}{\left(3z^{16/12}\right)(4)}-\frac{3z^{7/12}}{\left(4z^{9/12}\right)\left(3z^{7/12}\right)}$

$\displaystyle =\frac{-16-3z^{7/12}}{12z^{4/3}}$

$\displaystyle =-\frac{3z^{7/12}+16}{12z^{4/3}}$

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nevermind figured it out, thanks!

~sig

Quote:

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Originally Posted by sig

Yes that's right and this above is where I can get to. But here below I don't understand

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To add fractions together we need a common denominator. The least common multiple of 3 and 4 is 12, and we can treat $\displaystyle z^{4/3}$ as a common multiple of $\displaystyle z^{4/3}$ and $\displaystyle z^{3/4}$ because $\displaystyle \frac{z^{4/3}}{z^{3/4}}$$\displaystyle =z^{7/12}$$\displaystyle \Rightarrow z^{3/4}\cdot z^{7/12}$$\displaystyle =z^{4/3}$. So we shall make our common denominator $\displaystyle 12z^{4/3}$.

Now, how do we get this denominator in both fractions? By multiplying the numerator and denominator of the left fraction by 4 (because $\displaystyle 4\cdot3z^{4/3}=12z^{4/3}$) and by multiplying the numerator and denominator of the right fraction by $\displaystyle 3z^{7/12}$ (because, again, $\displaystyle 4z^{3/4}\cdot3z^{7/12}$$\displaystyle =12z^{4/3}$).

Quote:

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Originally Posted by sig

nevermind figured it out, thanks!

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I just finished typing this response as you posted, so I'll submit it anyway.

Thanks Reck,

I had been thinking combining fractional exponents were like combining fractions but I was totally wrong!

More than helpful, I'll be back sometime soon, I know it.

~sig