MHB Differentiate x^1/6 from first principles

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The discussion focuses on differentiating the function x^(1/6) using first principles. The user initially attempted to apply the difference of cubes and squares but encountered difficulties. The correct approach involves using the identity for a^6 - b^6, leading to the derivative formula: d(x^(1/6))/dx = (1/6)x^(-5/6). This method effectively simplifies the limit as h approaches 0, confirming the derivative's accuracy.

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Okay i got how to do x1/3 by using the difference of cubes and x1/2 by difference of squares but I am stuck on what todo with x1/6.

what i tried so far
F(x)= [ f(x+h) - f(x) ]/ h
a2= (x+h)1/6
b2= (x)1/6
so I eventually got to
[a6-b6]/h[a4+a2b2+b4

I used the difference of cubes with this or tried to atleast and it didnt work.

but i got the wrong answer by simplifying in the end, as i got 1/3x to the power of something
 
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omaarrr5 said:
Okay i got how to do x1/3 by using the difference of cubes and x1/2 by difference of squares but I am stuck on what todo with x1/6.

what i tried so far
F(x)= [ f(x+h) - f(x) ]/ h
a2= (x+h)1/6
b2= (x)1/6
so I eventually got to
[a6-b6]/h[a4+a2b2+b4

I used the difference of cubes with this or tried to atleast and it didnt work.

but i got the wrong answer by simplifying in the end, as i got 1/3x to the power of something

$\displaystyle \begin{align*} a^6 - b^6 &= \left( a^3 \right) ^2 - \left( b^3 \right) ^2 \\ &= \left( a^3 - b^3 \right) \left( a^3 + b^3 \right) \\ &= \left( a - b \right) \left( a^2 + a\,b + b^2 \right) \left( a^3 + b^3 \right) \end{align*}$

So in your case you can multiply top and bottom by:

$\displaystyle \begin{align*} \left[ \left( x + h \right) ^{\frac{1}{3}} + x^{\frac{1}{6}} \left( x + h \right) ^{\frac{1}{6}} + x^{\frac{1}{3}} \right] \left[ \left( x + h \right) ^{\frac{1}{2}} + x^{\frac{1}{2}} \right] \end{align*}$
 
You need to use the fact that
(a- b)(a^5+ a^4b+ a^3b^2+ a^2b^3+ ab^4+ b^5)= a^6- b^6

With a= (x+ h)^{1/6} and b= x^{1/6} that says
((x+h)^{1/6}- x^{1/6})((x+h)^{5/6}+ (x+h)^{4/6}x^{1/6}+ (x+h)^{3/6}x^{2/6}+ (x+h)^{2/6}x^{3/6}+ (x+h)^{1/6}x^{4/6}+ x^{5/6})= x+h- x= h

So that
\frac{(x+h)^{1/6}- x^{1/6}}{h}= \frac{1}{(x+h)^{5/6}+ (x+h)^{4/6}x^{1/6}+ (x+h)^{3/6}x^{2/6}+ (x+h)^{2/6}x^{3/6}+ (x+h)^{1/6}x^{4/6}+ x^{5/6}}

And, taking the limit as h goes to 0, on the right, every monomial becomes x^{5/6} and there are 6 of them so
\frac{dx^{1/6}}{dx}= \frac{1}{6x^{5/6}}= \frac{x^{-5/6}}{6}
 
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