Derivative of f(x) = x^{1/3}, the long way

[Pontifex]

Let:

\(\displaystyle f(x) = \root 3\of{x}\)

Using this form of the derivative, find \(\displaystyle f\prime(a)\): (Better way to prime that?)

\(\displaystyle \lim_{x\to a}\)\(\displaystyle {f(x) - f(a)} \over {x - a}\)

Thus:

\(\displaystyle \lim_{x\to a}\)\(\displaystyle {\root 3\of{x} - \root 3\of{a}} \over {x - a}\), Where \(\displaystyle a\neq0\)

My intuition is that \(\displaystyle {x - a}\) can be factored into something involving \(\displaystyle {\root 3\of{x} - \root 3\of{a}}\). Normally - if the numerator had an even exponent - I could use completing the square to wipe out the radicals, but I have yet to see a comparable strategy for the cube (which according to a quick google search, is surprisingly difficult, if geometrically tractable). My intuition is re-affirmed by knowing the derivative is \(\displaystyle {1/3}{x^{-2/3}}\), but darned if I can see a way to get there from here.

--Pontifex

 

[daon]

Your intuition is fine.

For all real numbers \(\displaystyle x\), \(\displaystyle (\sqrt[3]{x} )^3 = x\)

Use:

\(\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)\)

 

[Deleted member 4993]

Let:

\(\displaystyle f(x) = \root 3\of{x}\)

Using this form of the derivative, find \(\displaystyle f\prime(a)\): (Better way to prime that?)\(\displaystyle f'(a)\):

\(\displaystyle \lim_{x\to a}\)\(\displaystyle {f(x) - f(a)} \over {x - a}\)

Thus:

\(\displaystyle \lim_{x\to a}\)\(\displaystyle {\root 3\of{x} - \root 3\of{a}} \over {x - a}\), Where \(\displaystyle a\neq0\)

My intuition is that \(\displaystyle {x - a}\) can be factored into something involving \(\displaystyle {\root 3\of{x} - \root 3\of{a}}\). Correct Normally - if the numerator had an even exponent - I could use completing the square to wipe out the radicals, but I have yet to see a comparable strategy for the cube (which according to a quick google search, is surprisingly difficult, if geometrically tractable). My intuition is re-affirmed by knowing the derivative is \(\displaystyle {1/3}{x^{-2/3}}\), but darned if I can see a way to get there from here.

--Pontifex

\(\displaystyle \lim_{x\to a}\frac {\root 3\of{x} - \root 3\of{a}}{x - a}\),

\(\displaystyle =\lim_{x\to a}\frac {\root 3\of{x} - \root 3\of{a}}{(\root 3\of{x})^3 - (\root 3\of{a})^3}\)

\(\displaystyle = \lim_{x\to a}\frac{1}{[x^{\frac{2}{3}} + x^{\frac{1}{3}}\cdot a^{\frac{1}{3}} + a^{\frac{2}{3}}]}\)

and so on ....

 

[soroban]

Hello, Pontifex!

You are correct!
And this is what everyone is hinting about . . .

\(\displaystyle \text{Let: }\:f(x) \:=\: \sqrt[3]{x}\)

\(\displaystyle \text{Using: }\:\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\quad \text{find }f'(x)\)

\(\displaystyle \text{Since }\:a^3 - b^3 \:=\a-b)(a^2+ab + b^2)\)

. . \(\displaystyle \text{then: }\:x - a \:=\:\left(x^{\frac{1}{3}}\right)^3 - \left(a^{\frac{1}{3}}\right)^3 \;=\;\left(x^{\frac{1}{3}} - a^{\frac{1}{3}}\right)\left(x^\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}\right)\)


\(\displaystyle \text{We have: }\:\frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x-a}\)

\(\displaystyle \text{Multiply top and bottom by }\:x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}\)

. . \(\displaystyle \frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x - a}\cdot \frac{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}} {x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}} \;=\;\frac{x-a}{(x-a)\left(x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}\right)} \;=\;\frac{1}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}\)


\(\displaystyle \text{Then: }\;f'(x) \;=\;\lim_{x\to a}\left[\frac{1}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}\right]\)

. . \(\displaystyle \text{Got it?}\)

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^{1/3}, \ f'(x) \ = \ D_x[x^{1/3}] \ = \ \frac{1}{3x^{2/3}}.\)

\(\displaystyle f'(x) \ = \ lim \ \frac{f(x+h)-f(x)}{h} \ = \ lim\frac{(x+h)^{1/3}-x^{1/3}}{h} \ as \ h \ appproaches \ 0.\)

\(\displaystyle Now \ (a-b) \ = \ (a^{1/3}-b^{1/3})[a^{2/3}+(ab)^{1/3}+b^{2/3}]\)

\(\displaystyle If \ you're \ sincere, \ Pontifex, \ you \ should \ be \ able \ to \ take \ it \ from \ here.\)

 

[Pontifex]

Your intuition is fine.

For all real numbers \(\displaystyle x\), \(\displaystyle (\sqrt[3]{x} )^3 = x\)

Use:

\(\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)\)

Perfect Daon, thanks.

 

[Pontifex]

Let:

\(\displaystyle f(x) = \root 3\of{x}\)

Using this form of the derivative, find \(\displaystyle f\prime(a)\): (Better way to prime that?)\(\displaystyle f'(a)\):

Ah single quote gets turned into a prime, interesting:

\(\displaystyle f'(a)\)

This works also, probably more portable if the board here uses a special dialect of TeX specially designed for math:

\(\displaystyle f^\prime(a)\)

 

[Pontifex]

\(\displaystyle f(x) \ = \ x^{1/3}, \ f'(x) \ = \ D_x[x^{1/3}] \ = \ \frac{1}{3x^{2/3}}.\)

\(\displaystyle f'(x) \ = \ lim \ \frac{f(x+h)-f(x)}{h} \ = \ lim\frac{(x+h)^{1/3}-x^{1/3}}{h} \ as \ h \ appproaches \ 0.\)

\(\displaystyle Now \ (a-b) \ = \ (a^{1/3}-b^{1/3})[a^{2/3}+(ab)^{1/3}+b^{2/3}]\)

\(\displaystyle If \ you're \ sincere, \ Pontifex, \ you \ should \ be \ able \ to \ take \ it \ from \ here.\)

Easily, thanks. I just didn't know a cube could be factored like that. I'll have a start a cheat sheet for all of these things...

--Pontifex