Calc2.56

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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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