<s>[INST] 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 $x + h$:
$$
\text{First principle formula}: f(x) = \lim_{h \to 0} \frac{f(x + h) - f (x)}{h}\\\\
\text{determine}:f(x + h)\\
f(x) = (x)^\frac{1}{4}\\\\
f(x) = (\sqrt[4]{x})\\\\
f(x + h) = (\sqrt[4]{x+h})
$$
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!
 [/INST] 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 $h\to 0$,
$$\frac{f(x + h) - f (x)}{h}=\frac{1}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{1}{\sqrt{x+h}+\sqrt{x}}\to \frac{1}{2\sqrt[4]{x}}\cdot
\frac{1}{2\sqrt{x}}=\frac{1}{4x^{3/4}}.$$
</s>