# Math Help - derivative question

1. ## derivative question

Let f(x)= x1/3

If a does not = 0, use the eqaution f(a)= [f(x)-f(a)]/(x-a) to find f(a).

f`(a) = [x1/3-a1/3]/(x-a)

If the variables in the numerator were square roots rather than cube roots this would be much easier to solve. I'm not sure how to simplify this to find the derivative function. The answer in the back of the book is (1/3)a-2/3, but I can't get anywhere near this. Not sure how to even begin to break down the eqaution.

2. ## Re: derivative question

Yes, $a^2- b^2= (a- b)(a+ b)$ so, letting $a= x^{1/2}$ and $b= y^{1/2}$, we have $x- y= (x^{1/2}- y^{1/2})(x^{1/2}+ y^{1/2})$ so that $\frac{x^{1/2}- y^{1/2}}{x- y}= \frac{1}{x^{1/2}+ y^{1/2}}$.

Fortunately there is a similar, though a bit more complicated rule for third (and one-third) powers: $a^3- b^3= (a- b)(a^2+ ab+ b^2)$ so, letting $a= x^{1/3}$ and $b= y^{1/3}$,
$x- y= (x^{1/3}- y^{1/3})(x^{2/3}+ x^{1/3}y^{1/3}+ y^{2/3}$ so that $\frac{x^{1/3}- y^{1/3}}{x- y}= \frac{1}{x^{2/3}+ x^{1/3}y^{1/3}+ y^{2/3}}$.

3. ## Re: derivative question

Hello, cdbowman42!

$\text{Let }f(x) \,=\, x^{\frac{1}{3}}$

$\text{Use }\,\lim_{x\to a}\frac{f(x) - f(a)}{x-a}\,\text{ to find }f'(a).$

$\text{We have: }\:\frac{f(x) - f(a)}{x-a} \;=\;\frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x-a}$

$\text{Multiply by }\frac{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}\;\;\color{red}{**}$

. . $\frac{x^{\frac{1}{3}} - a^{\frac{1}{3}}}{x-a}\cdot \frac{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}} \;=\;\frac{x-a}{(x-a)(x^{\frac{2}{3}}+x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}})} \;=\;\frac{1}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}}$

$\text{Therefore: }\:f'(a) \;=\;\lim_{x\to a} \frac{1}{x^{\frac{2}{3}} + x^{\frac{1}{3}}a^{\frac{1}{3}} + a^{\frac{2}{3}}} \;=\;\frac{1}{a^{\frac{2}{3}} + a^{\frac{2}{3}} + a^{\frac{2}{3}}} \;=\;\frac{1}{3a^{\frac{2}{3}}}$

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**

Cube roots require a special "conjugate".

With fourth roots: . $\frac{x^{\frac{1}{4}} - a^{\frac{1}{4}}}{x-a}$

. . multiply by: . $\frac{x^{\frac{3}{4}} + x^{\frac{2}{4}}a^{\frac{1}{4}} + x^{\frac{1}{4}}a^{\frac{2}{4}} + a^{\frac{3}{4}}}{x^{\frac{3}{4}} + x^{\frac{2}{4}}a^{\frac{1}{4}} + x^{\frac{1}{4}}a^{\frac{2}{4}} + a^{\frac{3}{4}}}$

4. ## Re: derivative question

You could also do it as explained in attachment.