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 $x + h$:
$$ \text{First principle formula}: f(x) = \lim_{h \to 0} \frac{f(x + h) - f (x)}{h}\\\\ \text{determine}:f(x + h)\\ f(x) = (x)^\frac{1}{4}\\\\ f(x) = (\sqrt[4]{x})\\\\ f(x + h) = (\sqrt[4]{x+h}) $$
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!
$\endgroup$ 13 Answers
$\begingroup$Note that \begin{align*} f(x + h) - f (x)&=\left(\sqrt[4]{x+h}-\sqrt[4]{x}\right)\cdot \frac{\sqrt[4]{x+h}+\sqrt[4]{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{(x+h)-x}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot \frac{1}{\sqrt{x+h}+\sqrt{x}}. \end{align*} Hence as $h\to 0$, $$\frac{f(x + h) - f (x)}{h}=\frac{1}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot \frac{1}{\sqrt{x+h}+\sqrt{x}}\to \frac{1}{2\sqrt[4]{x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{4x^{3/4}}.$$
$\endgroup$ 4 $\begingroup$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}
$\endgroup$ $\begingroup$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} $
$\endgroup$ 3
