# Using the 4 step process, i.e. [f(x+h) - f(x)] / h...to find the derivative of the cube root of x.

thilina-g | College Teacher | (Level 1) Educator

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The definition of the derivative of f(x),

`f'(x) = lim_(h-gt0)[f(x+h)-f(x)]/h`

f(x) = `x^(1/3)` then, `f(x+h) = (x+h)^(1/3)`

then from the definition,

`f'(x) = lim_(h-gt0)[(x+h)^(1/3)-x^(1/3)]/h`

now if we try to evaluate this limit straight away, we will get, 0/0 which is not the answer. So we have to modify this expression, actually the numerator.

Before that let me remind you the simple formula of substraction of two cubics.

`a^3 - b^3 = (a-b)(a^2+ab+b^2)`

so, we can use this property to modify our expression. For our case,

`a = (x+h)^(1/3)`

`b = x^(1/3)`

I am going to multiply both numerator and denominator by ((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)).

`f'(x) = lim_(h-gt0)(((x+h)^(1/3)-x^(1/3))*((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))/(h*((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))`

`f'(x) = lim_(h-gt0)(((x+h)-x))/(h*((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))`

`f'(x) = lim_(h-gt0)h/(h*((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))`

since` h-gt0, h!=0`

so,

`f'(x) = lim_(h-gt0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3))`

`f'(x) = lim_(h-gt0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)) = 1/ ((x+0)^(2/3)+(x+0)^(1/3)x^(1/3)+x^(2/3))`

`f'(x) = lim_(h-gt0)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))`

`f'(x) = lim_(h-gt0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)) = 1/ (x^(2/3)+x^(2/3)+x^(2/3))`

`f'(x) = lim_(h-gt0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)) = 1/ (3x^(2/3))`

`f'(x) = lim_(h-gt0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)) = 1/ 3* x^(-2/3)`

The derivative of `root(3)(x)` is `1/3 x^(-2/3)`