Differentiate x^1/6 from first principles

[omaarrr5]

Okay i got how to do x^1/3 by using the difference of cubes and x^1/2 by difference of squares but I am stuck on what todo with x^1/6.

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

 

[Prove It]

Okay i got how to do x^1/3 by using the difference of cubes and x^1/2 by difference of squares but I am stuck on what todo with x^1/6.

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:

$\displaystyle \begin{align*} \left[ \left( x + h \right) ^{\frac{1}{3}} + x^{\frac{1}{6}} \left( x + h \right) ^{\frac{1}{6}} + x^{\frac{1}{3}} \right] \left[ \left( x + h \right) ^{\frac{1}{2}} + x^{\frac{1}{2}} \right] \end{align*}$

 

[HallsofIvy]

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
$$\frac{dx^{1/6}}{dx}= \frac{1}{6x^{5/6}}= \frac{x^{-5/6}}{6}$$