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OpenStudy (anonymous):
find the derivative of -6/x when x=12..help please...
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OpenStudy (amistre64):
you are given a process to work out .. what have you got?
OpenStudy (anonymous):
ok,
(-6/(x+h)-(6/x))/h
OpenStudy (amistre64):
youve dropped a negative
OpenStudy (anonymous):
ah so
((-6/(x+h)-(-6/x))/h
OpenStudy (anonymous):
common denom. so:-6x^2-6xh+6x^2+6xh
now that's the part I'm confused about
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OpenStudy (amistre64):
(-6/(x+h)--6/x)/h
(-6x/x(x+h)+ 6(x+h)/x(x+h))/h
( (-6x+6x+6h)/x(x+h) )/h
( (6h)/x(x+h) )/h
6h/hx(x+h)
6/x(x+h)
OpenStudy (amistre64):
\[\lim_0~\dfrac{\dfrac{-6}{x+h}-\dfrac{-6}{x}}{h}\]
\[\lim_0~\dfrac{\dfrac{-6x+6(x+h)}{x(x+h)}}{h}\]
\[\lim_0~\dfrac{-6x+6(x+h){}}{hx(x+h)}\]
\[\lim_0~\dfrac{\cancel{-6x+6x}+6h{}}{hx(x+h)}\]
\[\lim_0~\dfrac{6\cancel{h}{}}{\cancel{h}x(x+h)}\to \frac{6}{x(x+0)}\]
OpenStudy (amistre64):
let x=12 :)
OpenStudy (anonymous):
I see it, Thank you very much :DD
OpenStudy (amistre64):
youre welcome
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OpenStudy (anonymous):
do you mind helping me on another derivative question?
OpenStudy (anonymous):
f(x)=5x+9 x=2
OpenStudy (anonymous):
@amistre64
OpenStudy (anonymous):
i got
(5(x+h)+9)-(5x+9)/h
5x+5h+9-5x-9/h
OpenStudy (amistre64):
the slope of a line is constant ... regardless of any point in the line
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OpenStudy (amistre64):
you did fine so far now subtract what gets subtracted
OpenStudy (anonymous):
thank you, so it would be just be 5 then..
OpenStudy (amistre64):
yes
OpenStudy (anonymous):
thank you soo much :DD i think i got this now
OpenStudy (amistre64):
practice helps ;)
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