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This derivative is one of the worked examples in the MIT Open Course Ware cited below. All I am really doing here is reworking in a different notation. In Geometric Interpretation of Derivative I looked at this function and found the derivative at the particular point P = (2/3, 3/2). Here we find the derivative for the general x wherever f(x) is defined.

This is really a special case of the *Powers Rule*:

{$$ f'(x^n) = {d \over dx} x^n = nx^{n-1} $$}

because {$ 1/x = x^{-1} $}.

The definition of derivative I will use is:

{$$ f^\prime (x) = \lim_{ h \rightarrow0} {{f(x + h) -f(x)} \over { h}} $$}

which differs only in notation from other definitions.

Plug into the definition:

{$$ \begin{align} f^\prime (x) &= \lim_{ h \rightarrow0} {{f(x + h) -f(x)} \over { h}} \cr &= \lim_{ h \rightarrow0} {{ 1/(x + h) - 1/x} \over { h}} \end{align} $$}

As always in derivatives the limit cannot be solved by substitution because the denominator would become 0.

The next step is to express the fractions in the denominator with a common denominator:

{$$ \begin{align} f^\prime (x)&= \lim_{ h \rightarrow0} {{ 1/(x + h) - 1/x} \over { h}} \cr &= \lim_{ h \rightarrow0} {{ (x - (x +h))/(x(x + h)) } \over { h}} \cr &= \lim_{ h \rightarrow0} {{ -h /(x^2 + xh) } \over { h}} \cr &= \lim_{ h \rightarrow0} {{ h (-1 /(x^2 + xh)) } \over { h}} \end{align} $$}

Now h in the denominator cancels h in the numerator, and the limit can be solved by substitution.

{$$ \begin{align} f^\prime (x)&= \lim_{ h \rightarrow0} {{ h (-1 /(x^2 + xh)) } \over { h}} \cr &= \lim_{ h \rightarrow0} {{ -1 } \over {x^2 + xh }} \cr &= \frac{ -1 }{x^2 + x(0) } \cr \therefore f^\prime (x) &= -1 \left( \frac{ 1 }{x^2 } \right) = -1x^{-2} \end{align} $$}

This of course is undefined when x=0, but so is the original function.

*Sources:*

- MIT OpenCourseWare http://ocw.mit.edu "18.01SC Single Variable Calculus: Fall 2010" Creative Commons License Terms. Session 2: Examples of Derivatives | Part A: Definition and Basic Rules | 1. Differentiation | Single Variable Calculus | Mathematics | MIT OpenCourseWare
- Example 1. f(x) = 1/x - MIT18_01SCF10_Ses2a.pdf

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