# Calc2.56

$f(x)=\sqrt{x}\,$

$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \to 0}\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$

Here, we will multiply both the numerator and the denominator of the fraction by $\sqrt{x+\Delta x}+\sqrt{x}$ to cancel out the square roots in the numerator. The result is

$f'(x)=\lim_{\Delta x \to 0}\frac{x+\Delta x-x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\lim_{\Delta x \to 0}\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\lim_{\Delta x \to 0}\frac{1}{\sqrt{x+\Delta x}+\sqrt{x}}$

At this point, plugging in 0 for Δx presents no problems. Doing so gives

$f'(x)=\frac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2\sqrt{x}}$

Thus, the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.

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