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.
$$\begin{align}f'(x)&amp;=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\
&amp;=\lim_{h\to0}\frac{\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt x}}{h}\\
&amp;=\lim_{h\to0}\frac{7\sqrt x-7\sqrt{x+h}}{h\sqrt{x(x+h)}}\\
&amp;=7\lim_{h\to0}\frac{\sqrt x-\sqrt{x+h}}{h\sqrt{x(x+h)}}\cdot\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}\\
&amp;=7\lim_{h\to0}\frac{-h}{h\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}\\
&amp;=7\lim_{h\to0}\frac{-1}{\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}\\
&amp;=-\frac{7}{2x^\frac{3}{2}}\end{align}$$
Hint:


Use binomial series expansion of $(x+h)^{\frac{-1}{2}}$