Note that x^2 - 2x + 1 = (x - 1)^2. Then:
f(x) = (x^2 - 2x + 1)^(1/3) = [(x - 1)^2]^(1/3) = (x - 1)^(2/3).
Differentiating using the Chain and Power Rules yields:
f'(x) = (2/3)(x - 1)^(-1/3)
f''(x) = (2/3)(-1/3)(x - 1)^(-4/3) = -2/[9(x - 1)^(4/3)].
I hope this helps!
f (x) = [ x ² - 2x + 1 ]^(1/3)
f `(x) = (1/3) [ x ² - 2x + 1 ]^(-1/3) ( 2x - 2 )
f " (x) is then given by :-
(-1/9)[x ² - 2x + 1]^(-4/3)(2x - 2) ² + (2/3)[x ² - 2x + 1]^(-1/3)
[ x ² - 2x + 1]^(-4/3) [ (-1/9) (2x - 2) ² + (2/3) (x ² - 2x + 1) ]
[ (-1/9) (2x - 2) ² + (2/3) (x ² - 2x + 1) ]
-----------------------------------------------------
[ x ² - 2x + 1 ]^(4/3)
[ (-1) (2x - 2) ² + (6) (x ² - 2x + 1) ]
9 [ x ² - 2x + 1 ]^(4/3)
[ (-1) (4x ² - 8x + 4) + (6) (x ² - 2x + 1) ]
2 x ² - 4x + 2
------------------------
2 ( x ² - 2x + 1 )
-----------------------
2 ( x - 1 ) ²
-------------------
9 ( x - 1 )^(8/3)
2
--------------------
9 ( x - 1 )^(2/3)
f(x) = (x^2-2x+1)^(1/3)
f'(x)=1/3(x^2-2x+1)^(-2/3) *(2x-2) ---using the chain rule.
f''(x) = -2/9(x^2-2x+1)^(-5/3)*(2x-2) + 1/3(x^2-2x+1)^(-2/3) *2 ---using the chain and product rules
f'(x) = 1/3 (2x - 2)
f''(x) = 2/3
f(x)=[x^2-2x+1]^1/3=>
.....=(x-1)^2/3=>
f '(x)=(2/3)(x-1)^(-1/3)=>
f"(x)=(-2/9)(x-1)^(-4/3)=>
.......=-2/[9(x-1)(x-1)^1/3]
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Answers & Comments
Verified answer
Note that x^2 - 2x + 1 = (x - 1)^2. Then:
f(x) = (x^2 - 2x + 1)^(1/3) = [(x - 1)^2]^(1/3) = (x - 1)^(2/3).
Differentiating using the Chain and Power Rules yields:
f'(x) = (2/3)(x - 1)^(-1/3)
f''(x) = (2/3)(-1/3)(x - 1)^(-4/3) = -2/[9(x - 1)^(4/3)].
I hope this helps!
f (x) = [ x ² - 2x + 1 ]^(1/3)
f `(x) = (1/3) [ x ² - 2x + 1 ]^(-1/3) ( 2x - 2 )
f " (x) is then given by :-
(-1/9)[x ² - 2x + 1]^(-4/3)(2x - 2) ² + (2/3)[x ² - 2x + 1]^(-1/3)
[ x ² - 2x + 1]^(-4/3) [ (-1/9) (2x - 2) ² + (2/3) (x ² - 2x + 1) ]
[ (-1/9) (2x - 2) ² + (2/3) (x ² - 2x + 1) ]
-----------------------------------------------------
[ x ² - 2x + 1 ]^(4/3)
[ (-1) (2x - 2) ² + (6) (x ² - 2x + 1) ]
-----------------------------------------------------
9 [ x ² - 2x + 1 ]^(4/3)
[ (-1) (4x ² - 8x + 4) + (6) (x ² - 2x + 1) ]
-----------------------------------------------------
9 [ x ² - 2x + 1 ]^(4/3)
2 x ² - 4x + 2
------------------------
9 [ x ² - 2x + 1 ]^(4/3)
2 ( x ² - 2x + 1 )
-----------------------
9 [ x ² - 2x + 1 ]^(4/3)
2 ( x - 1 ) ²
-------------------
9 ( x - 1 )^(8/3)
2
--------------------
9 ( x - 1 )^(2/3)
f(x) = (x^2-2x+1)^(1/3)
f'(x)=1/3(x^2-2x+1)^(-2/3) *(2x-2) ---using the chain rule.
f''(x) = -2/9(x^2-2x+1)^(-5/3)*(2x-2) + 1/3(x^2-2x+1)^(-2/3) *2 ---using the chain and product rules
f'(x) = 1/3 (2x - 2)
f''(x) = 2/3
f(x)=[x^2-2x+1]^1/3=>
.....=(x-1)^2/3=>
f '(x)=(2/3)(x-1)^(-1/3)=>
f"(x)=(-2/9)(x-1)^(-4/3)=>
.......=-2/[9(x-1)(x-1)^1/3]