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The derivative of f(x) = `|x|^3` has to be determined at x = 0.
The absolute value of a number x is equal to x for `x>= 0` and it is equal to -x for x <0.
Now the function `f(x) = |x|^3` is equivalent to f(x) = x^3 for `x >= 0` and for x < 0 it is equivalent to `f(x) = -x^3` .
From how `|x|^3` has been defined, for values of `x >= 0` , f'(x) = 3x^2 and for values of x<0 the derivative f'(x) = -3x^2.
But the problem requires the value of f'(x) at x = 0. At x = 0, 3x^2 = 0 and -3x^2 is also equal to 0. The value of f'(x) is 0 both when x approaches 0 from values greater than 0 as well as when x approaches 0 from values lesser than 0.
As a result, the derivative of `f(x) = |x|^3` at x = 0 is 0.
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