Question

Can someone help me get this derivative. I did it, but I keep getting stuck.

Answer 1

\[x^2+3e^x/2e^x-x\]

Answer 2

I get \[4e^x-x^2-2xe^x/(2e^x-x)^2\]

Answer 3

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

Answer 4

yes

Answer 5

its like that ?

Answer 6

ohk...
you may use product rule or quotient rule.

wats ur fav ? :)

Answer 7

I'm using quotient

Answer 8

\(\huge (\frac{f}{g})' = \frac{gf' - g'f}{g^2}\)

Answer 9

yes

Answer 10

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


\(\huge \frac{(2e^x-x)(x^2+3e^x)' - (2e^x-x)'(x^2+3e^x)}{(2e^x-x)^2}\)

Answer 11

yes

Answer 12

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


\(\huge \frac{(2e^x-x)(x^2+3e^x)' - (2e^x-x)'(x^2+3e^x)}{(2e^x-x)^2}\)


\(\huge \frac{(2e^x-x)(2x+3e^x) - (2e^x-1)(x^2+3e^x)}{(2e^x-x)^2}\)

Answer 13

yes

Answer 14

rest is just algebra :)

Answer 15

ok thats what I got, but i get 4e^x-x^2-2xe^x/(2e^x-x)^2

Answer 16

I checked on a website and it said my answer was wrong

Answer 17

am i wrong?

Answer 18

for the top i am getting

\(\large -2e^xx^2-x^2+xe^x+3e^x\)

Answer 19

\[x^a * x^b = x^{a+b}\]

\[\therefore e^x*e^x = e^{2x}\]

i see no e^{2x} terms

Answer 20

when I expand out the top I get \[4xe^x-2x^2-6e^{2x} -3xe^x-2x^2e^x-6e^2x-x^2-3e^x\]

Answer 21

oh, I will try it again

Answer 22

e^2x cancels away right

Answer 23

yes

Answer 24

my bad

Answer 25

umm after i get \[xe^x-2x^2+6e ^{2x}-2x^2e^x-6e^{2x}+x^2+3e^x\]

Answer 26

looks good,
combine like terms

Answer 27

-2e^xx^2-x^2+xe^x+3e^x

Answer 28

i get that

Answer 29

thats right,
dont forget the denominator... :)

Answer 30

cant I factor anything out ?

Answer 31

when I put into this site http://www.wolframalpha.com/input/?i=dy%2Fdx+of+ln%28%28cos%28x^2%2B4%29%2Fcos%28x^2%2B9%29%29^5%29
I get a different answer

Answer 32

thats a different question ?

Answer 33

i see sin/cos/logs in ur link :o

Answer 34

sorry, if you replace it with this question it would give u a different answer

Answer 35

sorry i gave u the wrong link

Answer 36

here is the correct link :-

http://www.wolframalpha.com/input/?i=d%2Fdx+%28x%5E2%2B3e%5Ex%29%2F%282e%5Ex-x%29

Answer 37

yes, why do they take out the x and e?

Answer 38

its giving exact same answer :)
just u dint enter the d/dx thing correctly in ur old link...

Answer 39

oly wolfram knows that -_-

Answer 40

its looking bit compact/neat i guess like that..

Answer 41

oh ok. bc i thought my answer was wrong and i kept freaking out

Answer 42

ahh happens ;)

simple way to feed wolfram is to use tick notation

(function)'

Answer 43

enter below :-

( (x^2+3e^x)/(2e^x-x) )'

Answer 44

it interprets the derivative correctly

http://www.wolframalpha.com/input/?i=%28+%28x%5E2%2B3e%5Ex%29%2F%282e%5Ex-x%29+%29%27

Answer 45

oh ok. ugh

Answer 46

thank you for your help.

Answer 47

np :)