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
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
Hello, Hanga!
I don't think a "layman" could handle this one . . .
Differentiate: .$\displaystyle f(x) \:=\:x^{\frac{1}{6}}$ . . . . using the definition of a derivative (?)
We have: .$\displaystyle f(x+h) - f(x) \;= \;\frac{(x+h)^{\frac{1}{6}} - x^{\frac{1}{6}}}{1}$
Now we do something bizarre . . . multiply top and bottom by:
. . $\displaystyle (x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}$
and we get: .$\displaystyle \frac{(x + h) - x} {(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$
. . . . . . . $\displaystyle = \;\frac{h}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$
$\displaystyle \text{Divide by }h\!:\;\;\frac{1}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}}$
Take the limit:
$\displaystyle \lim_{h\to0}\left[\frac{1}{(x+h)^{\frac{5}{6}} + (x+h)^{\frac{4}{6}}x^{\frac{1}{6}} + (x+h)^{\frac{3}{6}}x^{\frac{2}{6}} + (x+h)^{\frac{2}{6}}x^{\frac{3}{6}} + (x+h)^{\frac{1}{6}}x^{\frac{4}{6}} + x^{\frac{5}{6}}} \right]$
. . $\displaystyle = \;\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}}} $ .$\displaystyle = \;\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: .$\displaystyle \boxed{f'(x) \;=\;\frac{1}{6x^{\frac{5}{6}}} }$