(:title Derivative of 1/x from Definition. :) (:div id="data" :) (:div2 width=200px class=picture style="text-align: center; margin-right: auto; margin-left: auto;":)%width=200px%Attach:Main/small1overx.svg[[<<]][- '''f(x) = 1/x'''-]%% (:div2end:) !! [[#contents]] Contents (:divend:) This derivative is [[http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/MIT18_01SCF10_Ses2a.pdf|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 [[MathCalculusSingleVariableGeometricInterpretationOfDerivative|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 {$1/x = x^{-1}$}. ![[#def]]Definition of the Derivative 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. ![[#sol]]Solution 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. (:div style="clear: both;":) (:divend:) (:div style="text-align:center":) (:ads @=0:) (:divend:) ----------- '''''Sources:''''' #MIT OpenCourseWare http://ocw.mit.edu "18.01SC Single Variable Calculus: Fall 2010" [[http://ocw.mit.edu/terms|Creative Commons License Terms]]. [[http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/|Session 2: Examples of Derivatives | Part A: Definition and Basic Rules | 1. Differentiation | Single Variable Calculus | Mathematics | MIT OpenCourseWare]] #[[http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/MIT18_01SCF10_Ses2a.pdf|Example 1. f(x) = 1/x - MIT18_01SCF10_Ses2a.pdf]] '''''Recommended:''''' '''Category:''' [[!Math]] [[!Calculua]]