Question

I have an aswer can someone check me please......
Find the derivative of b(x)= ((e^x)/(2x+3))

Answer 1

\[\frac{ e^x }{ 2x+3 }\]

Answer 2

wait minute

Answer 3

My answer that I got was\[\frac{ xe^x-2e^x }{ (2x+3)^2}\]

Answer 4

you can use wolfram to check your result
but it is incorrect.

Answer 5

how is it incorrect

Answer 6

how did you get it ?

I would write the problem as

e^x (2x+3)^(-1)
and use the product rule

Answer 7

\[\frac{ e ^{x}(2x+3)-2e ^{x} }{ (2x-3)^{2} }\]

Answer 8

ok I think I got it would the answer be \[b(x)=\frac{ -2xe^x+e^x }{ (2x+3)^2 }\]

Answer 9

the rule is
let Primarily=P ,Numerator=P , derivative=d
\[\frac{ (dN * P) - (dP * N)}{ P ^{2} }\]

Answer 10

closer.

here is one way to do it
d/dx ( e^x (2x+3)^(-1) )=

e^x d/dx (2x+3)^-1 + (2x+3)^-1 d/dx e^x

the d/dx (2x+3)^-1 = -(2x+3)^-2 * 2= -2/(2x+3)^2
and the first term is
-2e^x/(2x+3)^2

the 2nd term is e^x/(2x+3)= e^x(2x+3)/(2x+3)^2
\[ \frac{-2e^x + e^x(2x+3)}{(2x+3)^2} \]
or
\[ \frac{3e^x -2e^x + 2xe^x}{(2x+3)^2} \]
\[ \frac{e^x + 2xe^x}{(2x+3)^2} \]

Answer 11

ok but would this work\[b(x)=\frac{ 2xe^x+e^x }{ (2x+3)^2 }\]

Answer 12

sory Numerator=N
in my answer

Answer 13

yes, that is the same thing

Answer 14

awesome thanks for your help

Answer 15

right