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OCW Scholar - Single Variable Calculus 14 Online
OpenStudy (anonymous):

what is the derivative of f(x) = (a*x)/(b+x)?

OpenStudy (anonymous):

\[\frac{ bxa ^{x-1}+x ^{2}a ^{x-1}-a ^{x} }{ (b+x)^{2} }\]

OpenStudy (anonymous):

In this problem, a and b are constants, so the only variable is x. You have a quotient, so we have to apply the quotient rule:\[\frac{ d }{ dx }\frac{ u }{ v }=\frac{ vu'-uv' }{ v^2 }\]In this problem,\[u=ax, u'=a, v=b+x, v'=1\]Therefore we get \[\frac{ d }{ dx }\frac{ ax }{ b+x }=\frac{ (b+x)a-ax }{ (b+x)^2 }\]The numerator expands to ba+xa-ax which is equal to ba, so the answer is \[\frac{ ba }{ (b+x)^2 }\]The interesting point about this exercise is that the variable x disappears from the numerator, so that we have only the constant ba. That's not surprising when you consider that the original formula can be changed as follows, where the first step involves adding and subtracting the same number, so we're just adding zero:\[\frac{ ax }{ b+x }=\frac{ ax+ab-ab }{ b+x }=\frac{ a(b+x)-ab }{ b+x }=a-\frac{ ab }{ b+x }\]Because a is a constant and the derivative of a constant is zero, your answer is the same as what you would get if you simply find the derivative of -ab/(b+x).

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