rational exponents

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October 19th 2009, 12:44 PM
DBA

rational exponents

Hello, I try to understand a step in my solution manual, but I can not figure out what they did. Please help me with some steps they obvoiusly left out...

x^4/5 * 2(x-4) + (x-4)^2 * 4/5x^-1/5 =

1/5x^-1/5 *(x-4) *[5*x*2+(x-4)*4]

How do they get from line 1 to line 2? Thanks for any help
October 19th 2009, 02:28 PM
Defunkt

Quote:

Originally Posted by DBA

Hello, I try to understand a step in my solution manual, but I can not figure out what they did. Please help me with some steps they obvoiusly left out...

x^4/5 * 2(x-4) + (x-4)^2 * 4/5x^-1/5 =

1/5x^-1/5 *(x-4) *[5*x*2+(x-4)*4]

How do they get from line 1 to line 2? Thanks for any help

$x^{\frac{4}{5}}\cdot 2(x-4) + (x-4)^2 \cdot \frac{4}{5}x^{-\frac{1}{5}} = \frac{1}{5}\cdot 5 \cdot x \cdot x^{-\frac{1}{5}} \cdot 2(x-4) +$ $(x-4)(x-4) \cdot 4 \cdot \frac{1}{5} \cdot x^{-\frac{1}{5}}$

$= \frac{1}{5}\cdot x^{-\frac{1}{5}} \cdot (x-4) \cdot 5 \cdot x \cdot 2 + \frac{1}{5}\cdot x^{-\frac{1}{5}}\cdot (x-4)(x-4)\cdot 4 =$ $\frac{1}{5}x^{-\frac{1}{5}}(x-4)(5 \cdot x \cdot 2) + \frac{1}{5}x^{-\frac{1}{5}}(x-4)( 4(x-4) ) = \frac{1}{5}x^{-\frac{1}{5}}(x-4)(5\cdot x \cdot 2 + (x-4)4)$
October 19th 2009, 03:09 PM
Soroban

Hello, DBA!

Quote:

I cannot figure out what they did.

$x^{\frac{4}{5}}\cdot2(x-4) \:+\: (x-4)^2 \cdot\frac{4}{5}x^{-\frac{1}{5}}$

. . $=\;\frac{1}{5}x^{-\frac{1}{5}}(x-4)\cdot\bigg[5\cdot x\cdot2+(x-4)]\cdot4\bigg]$

How do they get from line 1 to line 2?

Is that the way they wrote it ? . . . How ugly!

Evidently, this is a derivative of a product.

$y \:=\:x^{\frac{4}{5}}(x-4)^2$

Then: . $\frac{dy}{dx} \;=\;x^{\frac{4}{5}}\cdot2(x-4) + (x-4)^2\cdot\frac{4}{5}x^{-\frac{1}{5}}$

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

Get a common denominator: . $\frac{2x^{\frac{4}{5}}(x-4)}{1}\cdot{\color{blue}\frac{5x^{\frac{1}{5}}}{5x ^{\frac{1}{5}}}} + \frac{4(x-4)^2}{5x^{\frac{1}{5}}}$ . $= \;\frac{10x(x-4)}{5x^{\frac{1}{5}}} + \frac{4(x-4)^2}{5x^{\frac{1}{5}}}$

. . $= \;\frac{10x(x-4) + 4(x-4)^2}{5x^{\frac{1}{5}}} \;=\;\frac{2(x-4)\bigg[5x + 2(x-4)\bigg]}{5x^{\frac{1}{5}}} \;=\;\boxed{\frac{2(x-4)(7x-8)}{5x^{\frac{1}{5}}}}$

which can be written: . $\frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)$
October 19th 2009, 05:23 PM
DBA

Thanks for your help, I understand it now :-)
November 30th 2009, 07:31 AM
Alienis Back

Sorry. Mistake