What is the derivative of [(x)^(5/3)] and [(x)^(4/3)] using the long method ?

Question

The "long method" that I'm saying is:
f(x) = [(x)^(5/3)]
f(x + h) = ?
f(x + h) - f(x) = ?
[f(x + h) - f(x)] / h = ?
limit of {[f(x + h) - f(x)] / h} as h approaches 0 = ?

f(x) = [(x)^(4/3)]
f(x + h) = ?
f(x + h) - f(x) = ?
[f(x + h) - f(x)] / h = ?
limit of {[f(x + h) - f(x)] / h} as h approaches 0 = ?

Answer 1

What is the derivative of [(x)^(5/3)] and [(x)^(4/3)] using the long method?

__________________________ __________________________

NOTE: BOTH proofs rely on the identity that

a - b ≡(a^⅓ - b^⅓)(a^(2/3) + a^⅓*b^⅓ + b^(2/3))

Note that this is only a special case of

x³ - y³ = (x - y)( x² + xy + y²) where x = a^⅓ and y = b^⅓

__________________________ __________________________

f(x) = (x)^(5/3)

f(x + h) = (x + h)^(5/3)

f(x + h) - f(x) = (x + h)^(5/3) - (x)^(5/3)

[f(x + h) - f(x)] / h = [(x + h)^(5/3) - (x)^(5/3)]/h

= [{(x + h)^(5)}^(1/3) - {(x)^(5)}^(1/3)]/h
* [({(x + h)^(5)}^(2/3) + {(x + h)^(5)}^(1/3)*{(x)^(5)}^(1/3) + {(x)^(5)}^(2/3)]/[({(x + h)^(5)}^(2/3) + {(x + h)^(5)}^(1/3)*{(x)^(5)}^(1/3) + {(x)^(5)}^(2/3)]

= [{(x + h)^(5)}^(3/3) - {(x)^(5)}^(3/3)]/[h*[{(x + h)^(5)}^(2/3) + {(x + h)^(5)}^(1/3)*{(x)^(5)}^(1/3) + {(x)^(5)}^(2/3)]]

= [x^5 + 5x^4 h + 10x³h² + 10x²h³ + 5xh^4 + h^5 - x^5}/[h*[{(x + h)^(5)}^(2/3) + {(x + h)^(5)}^(1/3)*{(x)^(5)}^(1/3) + {(x)^(5)}^(2/3)]]

= [h*(5x^4 + 10x³h + 10x²h² + 5xh³ + h^4)]
/[h*{(x + h)^(10/3) + (x + h)^(5/3) * x^(5/3) + x^(10/3)}]

= [5x^4 + 10x³h + 10x²h² + 5xh³ + h^4]
/[(x + h)^(10/3) + (x + h)^(5/3) * x^(5/3) + x^(10/3)]

limit h → 0 {[f(x + h) - f(x)] / h} =

limit h → 0 [5x^4 + 10x³h + 10x²h² + 5xh³ + h^4]
/[(x + h)^(10/3) + (x + h)^(5/3) * x^(5/3) + x^(10/3)]

= [5x^4]/[(x + h)^(10/3) + (x + h)^(5/3) * x^(5/3) + x^(10/3)]

= [5x^4]/[x^(10/3) + x^(5/3) * x^(5/3) + x^(10/3)]

= [5x^4]/[3x^(10/3)]

= 5/3 x^(4 - 10/3)

= 5/3 x^(2/3)

[= 5/3 x ^(5/3 - 1)]
__________________________ __________________________

f(x) = x^(4/3)

f(x + h) = (x + h)^(4/3)

f(x + h) - f(x) = (x + h)^(4/3) - x^(4/3)

[f(x + h) - f(x)] / h = [(x + h)^(4/3) - x^(4/3)]/h

= [{(x + h)^4}^(1/3) - {x^4^(1/3)]/h *
[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)] /


= [{(x + h)^4}^(3/3) - {x^4}^(3/3)] /
[h*[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]]

= [(x + h)^4 - x^4] /
[h*[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]

= [x^4 + 4x³h + 6x²h² + 4xh³ + h^4 - x^4] /
[h*[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]

= [h*(4x³ + 6x²h + 4h² + h³)] /
[h*[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]

= [4x³ + 6x²h + 4h² + h³] /
[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]


limit h → 0 {[f(x + h) - f(x)] / h} =

limit h → 0 [4x³ + 6x²h + 4h² + h³] /
[{(x + h)^4}^(2/3) + {(x + h)^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]

= 4x³/[{x^4}^(2/3) + {x^4}^(1/3) * {x^4}^(1/3) + {x^4}^(2/3)]

= 4x³/[x^(8/3) + x^(4/3) * x^(4/3) + x^(8/3)]

= 4x³/[3x^(4/3)]

= 4/3 x^(3 - 8/3)
= 4/3 x^(⅓)

[= 4/3 x^(4/3 - 1)]

Answer 2

f(x) = [(x)^(5/3)]
limit of {[f(x + h) - f(x)] / h} as h approaches 0
=limit of {[(x+h)^(5/3) - x^(5/3)] / h}
=limit of {x^(5/3) + (5/3)x^(2/3)h + O(h) - x^(5/3)] / h}
=limit of {(5/3)x^(2/3)h + O(h)] / h}
=(5/3)x^(2/3)

O(h) stands for the higher order of zero than h, which means O(h)/h approaches to zero when h approaches to zero.

In the same way, you can do the second problem.