Using first principles derive d/dx [x^(1/3)]`?

Question

Using the knowledge that (a^(1/3) - b^(1/3)).(a^(2/3) - a^(1/3).b^(1/3) + b^(2/3)) = a - b to help. Please explain how this was worked out. Cheers.

Answer 1

lim (h → 0) [f(x + h) - f(x)] / h

= lim (h → 0) [(x + h)^(1/3) - x^(1/3)] / h

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a^(1/3) - b^(1/3) = (a - b) / (a^(2/3) - a^(1/3).b^(1/3) + b^(2/3))

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= lim (h → 0) [(x + h - x) /((x + h)^(2/3) - (x + h)^(1/3).x^(1/3) + x^(2/3)) ] / h

= lim (h → 0) [h/((x + h)^(2/3) - (x + h)^(1/3).x^(1/3) + x^(2/3)) ] / h

= lim (h → 0) [1/((x + h)^(2/3) - (x + h)^(1/3).x^(1/3) + x^(2/3))

= 1/(x^(2/3) - x^(1/3).x^(1/3) + x^(2/3))

= 1/(x^(2/3) - x^(2/3) + x^(2/3))

= 1/x^(2/3)

= x^(-2/3)

I must have missed something as the answer should be 1/3*x^(-2/3)

I'll just go through it again

I found the mistake - you have a - where there should be a +

(a^(1/3) - b^(1/3)).(a^(2/3) + a^(1/3).b^(1/3) + b^(2/3)) = a - b

REDO of answer with mistake fixed

lim (h → 0) [f(x + h) - f(x)] / h

= lim (h → 0) [(x + h)^(1/3) - x^(1/3)] / h

..........................

a^(1/3) - b^(1/3) = (a - b) / (a^(2/3) + a^(1/3).b^(1/3) + b^(2/3))

..........................

= lim (h → 0) [(x + h - x) /((x + h)^(2/3) + (x + h)^(1/3).x^(1/3) + x^(2/3)) ] / h

= lim (h → 0) [h/((x + h)^(2/3) + (x + h)^(1/3).x^(1/3) + x^(2/3)) ] / h

= lim (h → 0) [1/((x + h)^(2/3) + (x + h)^(1/3).x^(1/3) + x^(2/3))

= 1/(x^(2/3) + x^(1/3).x^(1/3) + x^(2/3))

= 1/(x^(2/3) + x^(2/3) + x^(2/3))

= 1/[3x^(2/3)]

= 1/3 * x^(-2/3)