Explain Derviations of potensfunctions in laymens terms please

**Hanga** · April 16th 2008, 09:32 AM

Okay sitting here doing some homework that is using the simple derivaiton of potensfunctions... and I can't get my math to work here.
Please explain how you solve these one;

y = 6th square of x
y`= ?

Any help here is appreciated!
Thanks in advance

**Isomorphism** · April 16th 2008, 10:25 AM

Originally Posted by Hanga

Okay sitting here doing some homework that is using the simple derivaiton of potensfunctions... and I can't get my math to work here.
Please explain how you solve these one;

y = 6th square of x
y`= ?

Any help here is appreciated!
Thanks in advance

6th square of x??
Do you mean $\sqrt[6] {x}$ ?

**Hanga** · April 16th 2008, 10:33 AM

Originally Posted by Isomorphism

6th square of x??
Do you mean $\sqrt[6] {x}$ ?

Yeah indeed I do

I dunno how the Math commands on these forums work :P

anyways

whats the derivation of y = $\sqrt[6] {x}$

**Isomorphism** · April 16th 2008, 10:38 AM

Originally Posted by Hanga

Yeah indeed I do

I dunno how the Math commands on these forums work :P

anyways

whats the derivation of y = $\sqrt[6] {x}$

$y = \sqrt[6] {x}$

$y = {x}^{\frac16}$

$y' = \frac16{x}^{\frac16 - 1}$

$y' = \frac16{x}^{-\frac56}$

$y' = \frac16 \sqrt[6]{\frac1{x^5}}$

**Soroban** · April 16th 2008, 11:54 AM

Hello, Hanga!

I don't think a "layman" could handle this one . . .

Differentiate: . $f(x) \:=\:x^{\frac{1}{6}}$ . . . . using the definition of a derivative (?)

We have: . $f(x+h) - f(x) \;= \;\frac{(x+h)^{\frac{1}{6}} - x^{\frac{1}{6}}}{1}$

Now we do something bizarre . . . multiply top and bottom by:
. . $(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}$

and we get: . $\frac{(x + h) - x} {(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$

. . . . . . . $= \;\frac{h}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$

$\text{Divide by }h\!:\;\;\frac{1}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$

Take the limit:

$\lim_{h\to0}\left[\frac{1}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}} \right]$

. . $= \;\frac{1}{x^{\frac{5}{6}} + x^{\frac{4}{6}}x^{\frac{1}{6}} + x^{\frac{3}{6}}x^{\frac{2}{6}} + x^{\frac{2}{6}}x^{\frac{3}{6}} + x^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$ . $= \;\frac{1}{x^{\frac{5}{6}} + x^{\frac{5}{6}} + x^{\frac{5}{6}} + x^{\frac{5}{6}} + x^{\frac{5}{6}} + x^{\frac{5}{6}}}$

Therefore: . $\boxed{f'(x) \;=\;\frac{1}{6x^{\frac{5}{6}}} }$

**Moo** · April 16th 2008, 12:04 PM

Hello,

Why getting so complicated ? o.O
I think that people who still don't know general formulaes of derivation won't know they have to multiply by such a thing

**janvdl** · April 16th 2008, 12:31 PM

Originally Posted by Moo

Hello,

Why getting so complicated ? o.O
I think that people who still don't know general formulaes of derivation won't know they have to multiply by such a thing

Soroban did it using the definition of the derivative, not the power rule

(Okay now I'm seriously going to bed

)

**o_O** · April 16th 2008, 12:39 PM

Originally Posted by Soroban

. . . . . . . $= \;\frac{h}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$

$\text{Divide by }h\!:\;\;\frac{1}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$

Am I missing something? You divided h from the top but not from the bottom?

**Moo** · April 16th 2008, 12:44 PM

He divided by h : $f(x+h)-f(x)=h/...$

By taking the limit -> derivative

**o_O** · April 16th 2008, 12:48 PM

Ooh! He didn't divide by h yet. I see I see.

**Hanga** · April 16th 2008, 01:23 PM

Thanks guys!

Not only did you show me the correct answer, you also made me see how to prove that the calcutlation is correct.. hallelujah!

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