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)) ?

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Charles Anctil · Accepted Answer

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))#

Elijah Abram · Answer

undefined To prove that the derivative of x^(1/3) is 1/(3x^(2/3)), we can use the identity (x-y)(x^2+xy+y^2)=x^3-y^3.

Let's start by rewriting x^(1/3) as x^(1/3) = (x^(1/3))^1.

Now, we can rewrite the expression (x-y)(x^2+xy+y^2) as (x^(1/3) - y^(1/3))((x^(1/3))^2 + (x^(1/3))(y^(1/3)) + (y^(1/3))^2).

Expanding this expression, we get (x^(1/3))^3 - (y^(1/3))^3.

Since (x^(1/3))^3 is equal to x and (y^(1/3))^3 is equal to y, the expression simplifies to x - y.

Now, we can equate this to x^3 - y^3, which gives us x - y = x^3 - y^3.

To find the derivative of x^(1/3), we can differentiate both sides of the equation with respect to x.

The derivative of x - y with respect to x is 1, and the derivative of x^3 - y^3 with respect to x is 3x^2.

Therefore, we have 1 = 3x^2.

Simplifying this equation, we get x^2 = 1/3.

Taking the square root of both sides, we have x = ±√(1/3).

Since we are interested in the positive value of x, we have x = √(1/3).

Substituting this value back into the original expression x^(1/3), we get (√(1/3))^(1/3) = 1/(3√(1/3)).

Simplifying further, we have 1/(3√(1/3)) = 1/(3(1/√3)) = 1/(3/√3) = 1/(3/√3) * (√3/√3) = 1/(3√3/3) = 1/(√3) = √3/3.

Therefore, the derivative of x^(1/3) is 1/(3x^(2/3)).

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)) ?

Below is the clearer version of the question: Thank you.

Below is the clearer version of the question:

Thank you.