# Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$

So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ and know that applying

$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$

but am at a loss when trying to expand $(x+h)^\frac{4}{3}$

• You mean $(4/3) x^{1/3}$. Commented Aug 9, 2013 at 14:04
• Yes sorry, typo. Commented Aug 9, 2013 at 14:07

This is a little bit not fun, but will give your algebra a good workout. First note that $$(x+h)^{4/3}-x^{4/3}=\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{2/3}-x^{2/3}\right) .\tag{1}$$ and the right-hand side of (1) can be written as $$\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right) .\tag{2}$$ We have used the identity $a^2-b^2=(a+b)(a-b)$ a couple of times.

So we want to find $$\lim_{h\to 0}\frac{\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right)}{h}.\tag{3}$$

The first two terms in the numerator of (3) behave very nicely as $h\to 0$. We need to worry only about $$\frac{(x+h)^{1/3}-x^1/3}{h}.\tag{4}$$ Now we use the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$. Letting $a=(x+h)^{1/3}$ and $b=x^{1/3}$ we multiply top and bottom of (4) by $a^2+ab+b^2$. We get $$\frac{(x+h)-x}{h\left((x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3}\right)}.$$

Now everything is under control, you can put the pieces together.

Remark: There are many other ways to do the algebra. For example instead of the initial factoring that we did, you could immediately multiply top and bottom by $$(x+h)^{8/3}+(x+h)^{4/3}x^{4/3}+x^{8/3}.$$ Then on top we end up with $(x+h)^4-x^4$. We can expand this as usual and get a simpler-looking path to the answer.

You can use the generalized binomial coeficients to solve this problem, which gives you

$(x + h)^\frac{4}{3} = x^\frac{4}{3}(1 + \frac{4h}{3x} + \frac{4h^2}{18x^2} + \ldots)$

It will go on forever, but since you divide by $h$, most will tend to zero, and you'll be left with $x^\frac{4}{3} \frac{4}{3x} = \frac{4}{3}x^\frac{1}{3}$.

• Where do these come from? I think these are based on the Taylor expansion, which assumes the value of the derivative we seek. Commented Aug 9, 2013 at 14:19
• @RonGordon: why does $(1+h)^{\frac{4}{3}}=\sum_{k=0}^{\infty}\binom{\frac{4}{3}}{k}h^{k}$ assume the value of the derivative?
– Alex
Commented Aug 9, 2013 at 15:36
• @Alex: tell me how else you understand these coefficients for nonintegral values. I understand them as a sort of analytic continuation based on the value of the derivatives of $(1+x)^{\alpha}$ for nonintegral $alpha$. Commented Aug 9, 2013 at 15:59
• @RonGordon You can write $(1+x)^\alpha = e^{\alpha\log(1+x)}$, and using the derivative of $e^x$ and $\log(x)$ to prove the formula, but if you can go through the derivative of $e^x$ and $\log(x)$, you can also just compute the derivative via $x^{\frac{4}{3}} = e^{\frac{4}{3}\log(x)}$. Commented Aug 12, 2013 at 8:35

Let

$$D = \frac{(x+h)^{4/3}-x^{4/3}}{h}$$

Then

\begin{align}D^3 &= \frac{(x+h)^4-3 (x+h)^{8/3} x^{4/3} + 3 (x+h)^{4/3} x^{8/3} - x^4}{h^3}\\ &= \frac{4 x^3 h + 6 x^2 h^2 + 4 x h^3 + h^4 -3 (x+h)^{4/3} x^{4/3} \left [ (x+h)^{4/3}-x^{4/3}\right ]}{h^3}\\ &= \frac{4 x^3 + 6 x^2 h + 4 x h^2 +h^3}{h^2} - \frac{3 (x+h)^{4/3} x^{4/3}}{h^2} D\end{align}

Thus,

$$h^2 D^3 + 3 (x+h)^{4/3} x^{4/3} D = 4 x^3 + 6 x^2 h + 4 x h^2 +h^3$$

Taking the limit as $h \to 0$, (and assuming that $\bar{D}=\lim_{h \to 0} D$ remains finite as we do so), we see that

$$3 x^{8/3} \bar{D} = 4 x^3$$

or $\bar{D} = (4/3) x^{1/3}$ as was to be demonstrated.

Hint:

The expansion of $(x+h)^\frac{4}{3}$ is

$$h\sqrt[3]{x+h} + x\sqrt[3]{x+h}$$

So then you will have

$$\lim_{x \rightarrow \infty}\frac{h\sqrt[3]{x+h} + x\sqrt[3]{x+h} - x^{\frac{4}{3}}}{h}$$

Can you take it from there?