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bettyboop8904

  • 3 years ago

differentiate: y= (ae^{x} + b)/(ce^{x} + d)

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  1. bettyboop8904
    • 3 years ago
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    \[y=\frac{ ae ^{x} + b }{ ce ^{x} +d } \]

  2. CanadianAsian
    • 3 years ago
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    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. bettyboop8904
    • 3 years ago
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    Hmmm that's what I got but it says the answer is: \[\frac{ (ad-bc)e ^{x} }{ (ce ^{x}+d)^{2} }\]

  4. CanadianAsian
    • 3 years ago
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    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.

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