Answer:
[tex]\frac{2}{3a^{\frac{1}{3}}}[/tex]
Step-by-step explanation:
[tex]\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)[/tex].
[tex]f(x)=x^\frac{2}{3}[/tex]
Using power rule we get that [tex]f'(x)=\frac{2}{3}x^{-\frac{1}{3}}[/tex]
Evaluating this at [tex]x=a[/tex] gives us: [tex]f'(a)=\frac{2}{3}a^{-\frac{1}{3}}[/tex].
We could write without negative exponent giving us:
[tex]f'(a)=\frac{2}{3a^{\frac{1}{3}}}[/tex].
We could also go about it an algebraic way.
Notice the numerator is a difference of squares and can be factored as [tex](x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{1}{3}+a^\frac{1}{3})[/tex].
We need a factor in the numerator to be [tex]x-a[/tex] so we can get rid of the [tex]x-a[/tex] on bottom and then substitute [tex]a[/tex] for [tex]x[/tex].
Recall the difference of cubes formula:
[tex]p^3-q^3=(p-q)(p^2+pq+q^2)[/tex]
We are going to use this on the denominator:
[tex](x^\frac{1}{3})^3-(a^\frac{1}{3})^3[/tex]
[tex](x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})[/tex]
So that first factor there will actually cancel with a factor I mentioned for the numerator earlier.
Let's see it all together:
[tex]\lim_{x \rightarrow a}\frac{(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{1}{3}+a^\frac{1}{3})}{(x^\frac{1}{3}-a^\frac{1}{3})(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})}[/tex]
After the cancellation we have:
[tex]\lim_{x \rightarrow a}\frac{(x^\frac{1}{3}+a^\frac{1}{3})}{(x^\frac{2}{3}+x^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3})}[/tex]
Now we are ready to replace [tex]x[/tex] with [tex]a[/tex].
[tex]\frac{a^\frac{1}{3}+a^\frac{1}{3}}{a^\frac{2}{3}+a^\frac{1}{3}a^\frac{1}{3}+a^\frac{2}{3}}[/tex]
We have some like terms to combine:
[tex]\frac{2a^{\frac{1}{3}}}{a^\frac{2}{3}+a^{\frac{2}{3}}+a^\frac{2}{3}}[/tex]
[tex]\frac{2a^\frac{1}{3}}{3a^\frac{2}{3}}[/tex]
[tex]\frac{2}{3a^{\frac{2}{3}-\frac{1}{3}}}[/tex]
[tex]\frac{2}{3a^{\frac{1}{3}}}[/tex]
