I got this answer $$\frac{ae^r(b-1)}{(b + e^r)^2}.$$ But in the solution page, they put the answer $$\frac{abe^r}{ (b + e^r)^2}.$$
Which is the correct answer? There answer is correct if $d/dx(b) = 0$, I thought it's $1$.
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$h(r)=(ae^r) / (b + e^r)$? – Robert Z Mar 03 '19 at 18:35
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check now, I got typo mistake – Sports Highlights Mar 03 '19 at 18:38
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According to the quotient's rule, the derivative $h'(x)$ should be equal to $$\frac{(ae^r)'(b+e^r)-(ae^r)(b+e^r)'}{(b+e^r)^2}=\frac{ae^r(b+e^r)-(ae^r)(0+e^r)}{(b+e^r)^2}.$$ Can you take it from here? Did you find your mistake?
Robert Z
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why, $(b+e^r)'$ = $(0 + e^4)$ I thought it should be $(1 + e^4)$ – Sports Highlights Mar 03 '19 at 19:05
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An easy way to solve this uses $h-a=\frac{-ab}{b+e^r}\implies h^\prime=\frac{abe^r}{(b+e^r)^2}$. This approach replaces the quotient rule with the chain rule.
J.G.
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