Definition of the derivative to find the derivative of x^(1/3)

[Martinc31415]

Homework Statement

Use the definition of the derivative to find the derivative of x^(1/3)

Homework Equations

The Attempt at a Solution

[(x+h)^(1/3) - x^(1/3)]/h

I do not know where to go from here. If it were a square root I could conjugate.

 

[SammyS]

Homework Statement

Use the definition of the derivative to find the derivative of x^(1/3)

Homework Equations

The Attempt at a Solution

[(x+h)^(1/3) - x^(1/3)]/h

I do not know where to go from here. If it were a square root I could conjugate.

Hello Martinc31415. Welcome ton PF !

The difference of cubes can be factored, $a^3-b^3=(a-b)(a^2+ab+b^2)\,.$

So, suppose you have the difference of cube roots, $\displaystyle P^{1/3}-Q^{1/3}$. In this case, $\displaystyle P^{1/3} = a\,\ \text{ and }\ Q^{1/3} = b\,.$

Multiplying $\displaystyle \left(P^{1/3}-Q^{1/3}\right)$ by $\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)$ will give $\displaystyle \left(P^{1/3}\right)^3-\left(Q^{1/3}\right)^3=P-Q\,.$

Thus, $\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)$ acts as the "conjugate" for $\displaystyle \left(P^{1/3}-Q^{1/3}\right)\,.$

 

[Martinc31415]

oohh...

I would not have thought of that, ever!

Thanks a ton.