How does (x-y)(x^2+xy+y^2)=x^3-y^3 help to prove that derivative of x^(1/3) is 1/(3x^(2/3)) ?

Question

Below is the clearer version of the question:

Thank you.

2426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T17:58:45

See below.

Making #x= X^(1/3)# and #y = Y^(1/3)# we have

#(X-Y)/(X^(1/3)-Y^(1/3))=X^(2/3)+X^(1/3)Y^(1/3)+Y^(2/3)# or

#(X^(1/3)-Y^(1/3))/(X-Y)=1/(X^(2/3)+X^(1/3)Y^(1/3)+Y^(2/3))# now we know that

#lim_(Y->X)(X^(1/3)-Y^(1/3))/(X-Y)=d/(dX)X^(1/3) = lim_(Y->X)1/(X^(2/3)+X^(1/3)Y^(1/3)+Y^(2/3))# so

#d/(dX)X^(1/3)=1/(3X^(2/3)) = 1/(3 root(3)(X^2))#