Using the limit definition, how do you find the derivative of #f(x)=x^(1/3)#?

Question

William Abdalla · Accepted Answer

Use the fact that #a^3-b^3 = (a-b)(a^2+ab+b^2)#, so #a - b = (root3a-root3b)(root3a^2+root3(ab)+root3b^2)#

Therefore the conjugate of #root3(x+h) - root3x# is

#(root3(x+h)^2+root3(x+h)(root3x)+(root3x)^2)#.

The rest is analogous to finding the derivative of #x^(1/2)#.

I shall use the notation of the question with #x^(1/3)# rather than #root3(x)#

#lim_(hrarr0)((x+h)^(1/3)-x^(1/3))/h#

To save some space, let's do the algebra first, then find the limit.

#((x+h)^(1/3)-x^(1/3))/h = (((x+h)^(1/3)-x^(1/3)))/h * (((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))/(((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3))#

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

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

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

So, we have

#lim_(hrarr0)((x+h)^(1/3)-x^(1/3))/h = lim_(hrarr0)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))#

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

Bonus

It is also true that for positive integer #n#,

#a^n-b^n = (a-b)(a^(n-1)+a^(n-2)b+a^(n-3)b^2 + * * * +ab^(n-2)+b^(n-1))#.

This allow us to use the same general method for any #n^(th)# root.