# Computing derivatives with fractional exponents

I'm trying to compute the following derivative:

$$\text{Using first principles, differentiate}: f'(x) = (x)^\frac{1}{4}\\\\$$

I'm used to the functions being whole numbers or some simple algebra, i'm a little confused with what exactly to do when we're working with $(x)^\frac{1}{4}$.

Below is my attempt at determining $x + h$:

$$\text{First principle formula}: f(x) = \lim_{h \to 0} \frac{f(x + h) - f (x)}{h}\\\\ \text{determine}:f(x + h)\\ f(x) = (x)^\frac{1}{4}\\\\ f(x) = (\sqrt{x})\\\\ f(x + h) = (\sqrt{x+h})$$

This is where I get stuck, not sure how to determine it or substitute it into the formula and then simplify.

Any suggestions are welcomed, thanks!

• Hint: use $p^4-q^4=(p-q)(p^3+p^2q+pq^2+q^3)$. – Yves Daoust Sep 13 '17 at 16:19

Note that \begin{align*} f(x + h) - f (x)&=\left(\sqrt{x+h}-\sqrt{x}\right)\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{(x+h)-x}{\sqrt{x+h}+\sqrt{x}}\cdot \frac{1}{\sqrt{x+h}+\sqrt{x}}. \end{align*} Hence as $h\to 0$, $$\frac{f(x + h) - f (x)}{h}=\frac{1}{\sqrt{x+h}+\sqrt{x}}\cdot \frac{1}{\sqrt{x+h}+\sqrt{x}}\to \frac{1}{2\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{4x^{3/4}}.$$

• Thanks for your response Robert. This is quite a more complicated answer than I've done myself. I think I understand the gist of it, just need to go through it to fully grasp. – Rrr Sep 13 '17 at 16:23
• Hey Robert, I've had some time to look through your answer and was wondering if you could give me a hand understanding it. Are you able to step by step explain how you've removed the 4 sqrt in the first line and why you're multiplying it against itself over itself? Wouldn't that just be equivalent to 1 * 1. – Rrr Sep 15 '17 at 13:47
• Here we are using two times the identity $(a-b)(a+b)=(a^2-b^2)$. For example $(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})=\sqrt{x+h}-\sqrt{x}$. – Robert Z Sep 15 '17 at 14:14
• Thanks Robert, I have learnt that rule, apologies for being unclear but I was more confused by the division? In your comment you wrote how I would have approached the solution but I don't put it over itself (as you have done in the comment)? – Rrr Sep 15 '17 at 14:54

You have the special case of $a=1$ and $b=4$ of the derivative of $x^{a/b}$, which may be found using the geometric series:

$$\frac{x^n-h^n}{x-h}=x^{n-1}+x^{n-2}h+\dots+xh^{n-2}+h^{n-1}=\sum_{k=0}^{n-1}x^{n-k-1}h^k$$

We then proceed as follows:

\begin{align}\frac d{dx}x^{a/b}&=\lim_{h\to x}\frac{x^{a/b}-h^{a/b}}{x-h}\\\\&=\lim_{h\to x}\left[\vphantom{\frac{(x^{1/b})^b-(h^{1/b})^b}{x^{1/b}-h^{1/b}}}\frac{x^{a/b}-h^{a/b}}{x^{1/b}-h^{1/b}}\right]\div\left[\frac{(x^{1/b})^b-(h^{1/b})^b}{x^{1/b}-h^{1/b}}\right]\\\\&=\lim_{h^{1/b}\to x^{1/b}}\left[\vphantom{\frac{(x^{1/b})^b-(h^{1/b})^b}{x^{1/b}-h^{1/b}}}\frac{(x^{1/b})^a-(h^{1/b})^a}{x^{1/b}-h^{1/b}}\right]\div\left[\frac{(x^{1/b})^b-(h^{1/b})^b}{x^{1/b}-h^{1/b}}\right]\\\\&=\lim_{h^{1/b}\to x^{1/b}}\frac{\sum_{k=0}^{a-1}(x^{1/b})^{a-k-1}(h^{1/b})^k}{\sum_{k=0}^{b-1}(x^{1/b})^{b-k-1}(h^{1/b})^k}\\\\&=\frac{\sum_{k=0}^{a-1}(x^{1/b})^{a-k-1}(x^{1/b})^k}{\sum_{k=0}^{b-1}(x^{1/b})^{b-k-1}(x^{1/b})^k}\\\\&=\frac{\sum_{k=0}^{a-1}x^{(a-1)/b}}{\sum_{k=0}^{b-1}x^{(b-1)/b}}\\\\&=\frac abx^{(a-b)/b}\end{align}

Well, a bit handwavy but:

Not for any $n > 1$, $(x^n - y^n)= (x-y)(x^{n-1} + x^{n-2}y + .... + xy^{n-2}+ y^{n-1})$

So $h = (x+h) - x = [(x+h)^{\frac 14} - x^{\frac 14}][(x+h)^{\frac 34} + (x+h)^{\frac 12}x^{\frac 14} + (x+h)^{\frac 14}x^{\frac 12} + x^{\frac 34}]$

So for $h> 0$ then $\frac {(x+h)^{\frac 14} - x^{\frac 14}}{h} =\frac 1{(x+h)^{\frac 34} + (x+h)^{\frac 12}x^{\frac 14} + (x+h)^{\frac 14}x^{\frac 12} + x^{\frac 34}}$

take limit of that.

$\lim \frac {(x+h)^{\frac 14} - x^{\frac 14}}{h} =\lim \frac 1{(x+h)^{\frac 34} + (x+h)^{\frac 12}x^{\frac 14} + (x+h)^{\frac 14}x^{\frac 12} + x^{\frac 34}}=\frac 1 {4x^{\frac 34}}=\frac 14x^{-\frac 34}$

• I think the goal is to use the limit definition of the derivative. – Simply Beautiful Art Sep 13 '17 at 17:04
• I think the goal is to use the limit definition of the derivative. – Simply Beautiful Art Sep 13 '17 at 17:19
• Oh,....yeah. Okay. – fleablood Sep 13 '17 at 17:26