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 \Delta x 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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