Calc2.57

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f(x)={\frac  {1}{x}}\,

f'(x)=\lim _{{\Delta x\to 0}}{\frac  {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{{\Delta x\to 0}}{\frac  {{\frac  {1}{x+\Delta x}}-{\frac  {1}{x}}}{\Delta x}}=\lim _{{\Delta x\to 0}}{\frac  {{\frac  {x-(x+\Delta x)}{x(x+\Delta x)}}}{\Delta x}}=\lim _{{\Delta x\to 0}}{\frac  {{\frac  {-\Delta x}{x(x+\Delta x)}}}{\Delta x}}

=\lim _{{\Delta x\to 0}}{\frac  {-\Delta x}{x\Delta x(x+\Delta x)}}=\lim _{{\Delta x\to 0}}{\frac  {-1}{x(x+\Delta x)}}=-{\frac  {1}{x^{2}}}

Thus, we have shown that the derivative of {\frac  {1}{x}} is -1{\frac  {1}{x^{2}}}.


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