## anonymous 3 years ago differentiate: y= (ae^{x} + b)/(ce^{x} + d)

1. anonymous

$y=\frac{ ae ^{x} + b }{ ce ^{x} +d }$

2. anonymous

f(x) = ae^x + b, then f'(x) = ae^x g(x) = ce^x + d, then g'(x) = ce^x You have y = f(x)/g(x) The derivative of this is (f'(x)*g(x) - g'(x)*f(x))/g(x)^2 So ((ae^x)(ce^x + d) - (ce^x)(ae^x + b))/(ce^x)^2

3. anonymous

Hmmm that's what I got but it says the answer is: $\frac{ (ad-bc)e ^{x} }{ (ce ^{x}+d)^{2} }$

4. anonymous

if you multiply everything on the top out, you get ace^(2x) + ade^x - ace^(2x) - bce^x ===> ade^x - bce^x ==> (ad-bc)e^x I also mistyped something in my earlier answer, it should be (ce^x +d)^2 on the bottom. Sorry about that.