Question

differentiate e^(x+y) = ln(x+y)+3x+3y+3

Answer 1

use chain rule in both sides

Answer 2

first diffrentiate LHS can u do it?

Answer 3

Ok this what I got. Please see if it's correct. I'll do the left first. :)

Answer 4

yeah go ahead

Answer 5

\[e ^{x+y} y\prime\]

Answer 6

Is this correct?

Answer 7

what are you differentiating with respect to?

Answer 8

is this implicite?

Answer 9

almost theres a slight mistake look--
\[\frac{ d }{ dx }e^{x+y}=e^{x+y}*\frac{ d }{ dx}(x+y) =e^{x+y}*(1+y^{'})\]
ok?

Answer 10

got it? you missed that x

Answer 11

yes implicate

Answer 12

oh yeah, crap I forgot that one. Thanks. :)

Answer 13

now try RHS

Answer 14

\[\frac{ y\prime\ }{ x+y } +3 + 3y\prime\\] How's this?

Answer 15

Oh pellet wait wth

Answer 16

lol pellet i cant say sh*t here lolol

Answer 17

\[\frac{ y\prime\ }{ x+y } +3 + 3y\prime\\]

Answer 18

uhh yout latex is'nt working

Answer 19

Crap. can you try copying it to yours? maybe it'll show up.

Answer 20

ok i'm doing it ,check it with yours

Answer 21

\[\frac{ d }{ dx }[\ln(x+y)+3x+3y+3]=\frac{ 1 }{ x+y }\frac{ d }{ dx }(x+y)+3+3y^{'}+0\\=\frac{ 1 }{ x+y }*(1+y^{'})+3+3y^{'}\]
ok?

Answer 22

@1018

Answer 23

i forgot the 1 again. tsk. wait, cant it be 1+y(prime) / x +y ?

Answer 24

cant i put the 1+yprime as the numerator?

Answer 25

no you cant because u're diffrentiating (x+y) alone after differentiating it with log

Answer 26

oh. ok. cause in my notes, it says the formula is du/u is it wrong?

Answer 27

itd like
\[\frac{ d }{ dx }\ln(x^2+x^3)=\frac{ 1 }{ x^2+x^3 }*(2x+3x^2)\]
like this

Answer 28

yeah that, but in mine it's du/u. so is it wrong to be \[\frac{ 2x + 3x^2 }{ x^2 + x^3}\]

Answer 29

no its not wrong

Answer 30

oh. ok thanks for clarifying. so now, i got \[e ^{x+y} 1+y\prime\ = \frac{ 1+y\prime }{ x + y } + 3 + 3y\]

Answer 31

correction. that's 3yprime

Answer 32

yes thats correct :)

Answer 33

then what's next :)

Answer 34

then take things with y(prime) to a side take take y(prime) common

Answer 35

like this--
\[e^{x+y}+e^{x+y}*y^{'}=\frac{ 1 }{ x+y }+\frac{ y^{'} }{ x+y }+3+3y^{'}\\y^{'}[e^{x+y}-\frac{ 1 }{ x+y }-3]=\frac{ 1 }{ x+y }+3-e^{x+y}\]

Answer 36

got it?

Answer 37

wait, haha why are there two e in the lhs?

Answer 38

because we got
\[LHS=e^{x+y}(1+y^{'})=e^{x+y}+e^{x+y}*y^{'}\\just ~multiplied~them\]

Answer 39

oh yeah yeah. crap man i need to it i'm losing focus. thanks for the help. the final answer for this is -1, right?

Answer 40

*eat

Answer 41

i dont know you can calculate it easy algebra only :)

Answer 42

no our professor told us that it should be -1 lol

Answer 43

ok ok thanks appreciate your help!

Answer 44

yes it will be -1 just calculate it u'll get the answer

Answer 45

yw!!!