Question

Differentiate with e

Answer 1

\[y=\frac{ ae^x+b }{ ce^x+d }\]y'=?

Answer 2

Have we attempted the Quotient Rule? What has "e" to do with it?

Answer 3

\[\frac{ (ce^x+d)(ae^x+b)'-(ae^x+b)(ce^x+d)' }{ (ce^x+d)^2 }\]erm yeah, I did this. But i'm not sure how to differentiate with those a's and b's and e's.
E becaue of the e^x's in there.

Answer 4

I mean, are c,d,b, a all constants? do they become 1?

Answer 5

That';s good. \((ae^{x}+b)' = ae^{x}\) a, b, c, d, and e are just constants. e is a specific constant. The others are arbitrary.

Answer 6

\(\dfrac{d}{dx}a = 0\)

\(\dfrac{d}{dx}e = 0\)

\(\dfrac{d}{dx}a^{x} = a^{x}\ln(a)\)

\(\dfrac{d}{dx}e^{x} = e^{x}\ln(e) = e^{x}\)

Answer 7

hmm ok.
so in ae^x the a stays there because it is.."attached" to e?

Answer 8

It's just a constant and comes along for the ride.

Answer 9

\(\dfrac{d}{dx}ae^{x} = a\dfrac{d}{dx}e^{x} = ae^{x}\)

Answer 10

\(\dfrac{d}{dx}(ae)^{x} = (ae)^{x}\ln(ae) = (ae)^{x}(\ln(a)+1)\)

Answer 11

Try a few examples and get used to what is a constant and what isn't.

Answer 12

Ohh I see! That makes more sense written that way with the a before the d/dx

Answer 13

\[\frac{ e^x(ad-bc) }{ (c^ex+d)^2}\]=y'

Answer 14

oops that botton was meant to be
ce^x

Answer 15

I think you have it! Good work.

Very often, there will be some odd simplification with the denominator. I don't see any, this time. Keep your eyes out for it.

Answer 16

Thanks so much! :)

Answer 17

ah ok. In this one there was just a lot of multiplying in and then taking back out again xD