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

• May 20th 2012, 12:54 PM
sig
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 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
• May 20th 2012, 01:04 PM
Plato
Re: need help understanding d/dx(x^(1/2))
Quote:

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

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

Thus $\frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{ 2} x^{-\frac{1}{2}}$
• May 20th 2012, 01:15 PM
sig
Re: need help understanding d/dx(x^(1/2))
Quote:

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

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

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

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

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)
• May 20th 2012, 01:23 PM
Plato
Re: need help understanding d/dx(x^(1/2))
Quote:

Originally Posted by sig

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

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)

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

Recall that $\left(x^{-\frac{1}{2}}\right)=\frac{1}{x^{\frac{1}{2}}}$
• May 20th 2012, 01:26 PM
sig
Re: need help understanding d/dx(x^(1/2))
Quote:

Originally Posted by Plato

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

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

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
• May 20th 2012, 05:11 PM
sig
Re: need help understanding d/dx(x^(1/2))
Hello, I'm back.
After looking and working on this equation,

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

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
• May 20th 2012, 05:34 PM
Plato
Re: need help understanding d/dx(x^(1/2))
Quote:

Originally Posted by sig
Hello, I'm back.
After looking and working on this equation,

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

That is $\frac{d}{dz}\left(4z^{-\frac{1}{3}}\right)=\frac{-4}{3}z^{\frac{- 4}{3}}$
• May 20th 2012, 05:43 PM
sig
Re: need help understanding d/dx(x^(1/2))
Quote:

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

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?
• May 20th 2012, 05:51 PM
Reckoner
Re: need help understanding d/dx(x^(1/2))
Quote:

Originally Posted by sig
Hello, I'm back.
After looking and working on this equation,

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

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

We have

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

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

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

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

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

and, combining fractions,

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

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

$=-\frac{3z^{7/12}+16}{12z^{4/3}}$
• May 20th 2012, 05:54 PM
sig
Re: need help understanding d/dx(x^(1/2))
Quote:

$=-\frac4{3z^{16/12}}-\frac1{4z^{9/12}}$
Yes that's right and this above is where I can get to. But here below I don't understand

Quote:

and, combining fractions,

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

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

$=-\frac{3z^{7/12}+16}{12z^{4/3}}$
• May 20th 2012, 06:26 PM
sig
Re: need help understanding d/dx(x^(1/2))
nevermind figured it out, thanks!

~sig
• May 20th 2012, 06:32 PM
Reckoner
Re: need help understanding d/dx(x^(1/2))
Quote:

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

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

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

Quote:

Originally Posted by sig
nevermind figured it out, thanks!

I just finished typing this response as you posted, so I'll submit it anyway.
• May 20th 2012, 06:35 PM
sig
Re: need help understanding d/dx(x^(1/2))
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