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Mathematics 13 Online
OpenStudy (anonymous):

(d/dx)[(x+8)/(x^2+x+2)]=

OpenStudy (anonymous):

quotient again...

OpenStudy (anonymous):

if we get a equition=f(x)/g(x) the derivitive for the equition will be [f'(x)g(x)-f(x)g'(x)]/[g(x)]^2

OpenStudy (anonymous):

hmmmyou sure? I thought it was g(prime)(x)f(x)-...first...

OpenStudy (anonymous):

yep, I just check it after u asked

OpenStudy (amistre64):

(d/dx)(T(x)/B(x)) = [B(x)T'(x)-T(x)B'(x)]/[B(x)]^2 is how I remember it. T for top and B for bottom.

OpenStudy (anonymous):

edwardhugh is right\[d/dx(x+8)/(x^2+x+2)=\] \[[x^2+x+2-(x+8)(2x+9)]/(x^2+x+2)^2\] \[=(-x^2-16x-6)/(x^2+x+2)^2\]\[=-(x^2+16x+6)/(x^2+x+2)^2\] You can expand the denominator if you like, but this would be its simplified form

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