Question

find dy/dx using implicit differentiation. compare your answer with the result obtained by first solving for y as a function of x then taking the derivative. e^xy-y=4

Answer 1

\[e ^{xy-y} \]
?

Answer 2

its written as e^x y-y=4

Answer 3

its asking for implicit differentiation dy/dx= ?

Answer 4

as well as solving for y first then take the derivative, so y= ? and dy/dy=?

Answer 5

can't help you until I know what the problem is..

Answer 6

that is the problem the equation is e^x y-y=4

Answer 7

are you there?

Answer 8

I'll assume it's \[e ^{x} (y-y) = 4\]

Answer 9

e^x*0 =4
0=4
false.

Answer 10

IS IT
\[e ^{xy} - y =4\]
OR
\[e ^{xy- y} =4\]

Answer 11

it just says e^xy-y=4 that is exactly how it is written

Answer 12

cool.

Answer 13

thanks? no need to be rude here..

Answer 14

Yes. Excuse me for bullying you on the internet. Which version of the problem would you prefer to do?

Answer 15

huh?

Answer 16

\[e ^{xy} -y =4\]
or
\[e ^{xy -y} =4\]

Answer 17

is it not possible to solve a problem given e^xy-y=4, because thats how it is written online, is that incorrect or something?

Answer 18

It's entirely possible to solve it either way. Which one shall we solve?

Answer 19

lets go with the first one please

Answer 20

that tyou wrote

Answer 21

Excellent choice.

(e^u) ' = (e^u)*du
(e^(xy)) ' = (e^xy)*(xy' +y)

Answer 22

(e^xy)*(xy' +y) - y' =0

Answer 23

group terms with y' on one side of the equation, factor out y', divide both sides by the factor on y'

Answer 24

so I can subtract thewhole term e^xy from the other side?

Answer 25

I don't know what you mean. You can do anything that the rules of algebra allow.

Answer 26

is this fine? (xy'+y)-y'=-e^xy

Answer 27

no.

Answer 28

I thought we had to isolate the y' so so far I moved over the e^xy term

Answer 29

4*2 = 8
-4 -4
2 = 4

Answer 30

seem reasonable?

Answer 31

no

Answer 32

(e^xy)(xy'+y)=y'

Answer 33

?

Answer 34

group terms with y' on one side of the equation

Answer 35

e^xy=-xy'+y-y' ?

Answer 36

?

Answer 37

can you please just help me..

Answer 38

Perhaps it is \(e^xy-y = 4\)?

Answer 39

thats the problem how it is exactly written

Answer 40

I need to find the implicit differentiation of it, do you know how to do that?

Answer 41

rolypoly

Answer 42

Do you know who to differentiate \(e^xy\)?

Answer 43

*how

Answer 44

oh not for this problem, e^x would just be e^x but theres nothing to take the derivative of the y's..

Answer 45

You need to apply product rule when differentiating that guy. Perhaps you would like to try again.

Answer 46

wouldnt it just be e^xy dy/dx?

Answer 47

Not really.
Product rule: uv' + u'v

Answer 48

In this case, u = e^x , v=y
v' = y'
Can you try again?

Answer 49

e^x dy/dx+e^x(y)

Answer 50

Yup. \[e^xy-y=4\]
Diff. both sides w.r.t. x
What do you get?

Answer 51

e^x dy/dx+e^x(y)-y=4 what does w.r.t. x mean?

Answer 52

so do i have to get the e^x dy/dx separate?

Answer 53

w.r.t. = with respect to
e^x dy/dx+e^x(y) <- right
-y=4 <- not right.

Answer 54

And not yet, you can't separate them at the moment, until you get this step correct.

Answer 55

correct

Answer 56

so i caan subtract y from each side first

Answer 57

No!
Differentiate both sides with respect to x means that you have to differentiate EVERY TERMS on both sides with respect to x.

Answer 58

would it be e^x dy/dx+e^x(y)-y dy/dx=4?

Answer 59

No, -y dy/dx and are not right.

Answer 60

dy/dx = dy/dx, nothing more
d/dx (constant) = 0, as always.

Answer 61

oh ok sorry the y does not get dy/dx bc it is not being differntiated so its e^x dy/dx+e^x(y)=0

Answer 62

No...
d/dx (y) = dy/dx = y'
Similarly, d/dx (p) = dp/dx = p'

Answer 63

uhh im confused :/

Answer 64

can you please guide me im confused and i dont have much time left for this question :/

Answer 65

For the question find dy/dx of the equation y = 2x, how do you find it?

Answer 66

2x*x ?

Answer 67

No.

Answer 68

i think for my actualy problem is it y/e^xy

Answer 69

y=e^x

Answer 70

No.

Answer 71

what am i doing wrong :/

Answer 72

How do you get y=e^x? What is y/e^xy?

Answer 73

i dont know im just taking a guess..

Answer 74

my times running out :/

Answer 75

e^xy(xy)'-y'=0 ?

Answer 76

Almost correct!
For the part e^xy(xy)', it's not correct, but the -y'=0 part is correct.

Answer 77

e^x dy/dx+e^x(y)-y'=0

Answer 78

Yes. That's correct.
Now, dy/dx can be rewrite as y'
Make y' as the subject, can you do it?

Answer 79

so you mean it looks like e^x y'+e^x(y)-y'=0

Answer 80

and isoloate the y'

Answer 81

Yes!

Answer 82

:D

Answer 83

so can i subtract the e^xy'+e^x(y) all of that?

Answer 84

Subtract? why?

Answer 85

to the other side to get y' by itself

Answer 86

e^x y'+e^x(y)-y'=0
Rearrange it.
e^x y'-y'+e^x(y)=0
Subtract e^x (y) from both sides, what do you get?

Answer 87

e^xy'-y'=-e^x(y)

Answer 88

now subtract e^x to get y'-y' separate?

Answer 89

No.
Then, factorize the left part, what do you get?

Answer 90

y'(e^x-y')=-e^x(y)

Answer 91

No. Try again..

Answer 92

idk what to do with the other y' bc then it will be y'^2

Answer 93

When you expand y'(e^x-y'), you won't get (e^x y'-y') , so you did the factorization wrong.

Answer 94

i know i dont know how to do it

Answer 95

Take out y' as the common, what's the terms left?

Answer 96

e^x and y'