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Mathematics 18 Online
OpenStudy (anonymous):

find the derivative by definition y=1/x

OpenStudy (anonymous):

-1/x^2

OpenStudy (anonymous):

derivative (x)^-1 is -1 x^-2 you can rewrite -1/x^2

OpenStudy (anonymous):

by definition, do you mean taking the limit of the difference quotient?

OpenStudy (anonymous):

yeah by definiton means taking limit

OpenStudy (anonymous):

\[\lim_{h\to 0} \, \frac{\frac{1}{h+x}-\frac{1}{x}}{h}\]

OpenStudy (anonymous):

\[\lim_{h\to 0} \, \frac{-\frac{h}{x (h+x)}}{h}\]

OpenStudy (anonymous):

\[\lim_{h\to 0} \, {-\frac{h}{x h(h+x)}}\] \[\lim_{1\to 0} \, {-\frac{1}{x (h+x)}}\]

OpenStudy (anonymous):

as h goes to 0 , -1/x^2

OpenStudy (anonymous):

can you please solve it with y+lemda y rule

OpenStudy (anonymous):

I am not sure what you mean?

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