<s>[INST] Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the problem, I would be eternally grateful.
 [/INST] $$\begin{align*}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac7h\cdot \left( \frac1{\sqrt{x+h}} -\frac1{\sqrt{x}} \right)\\ 
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{x-(x+h)}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\right)\\
&=\lim_{h\to 0}\frac{-7}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\\
&=\frac{-7}{x\cdot 2\sqrt x}
\end{align*}$$
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