Question

calculate the derivative of y with respect to x.
(x^3)y+3x(y^3)= x + y

Answer 1

Do you know about the product rule of derivative??

Answer 2

Yes i do and applied it, but for some reason the answer i've been getting is not correct.

Answer 3

Can you show me the entire steps so that I can fine the mistake where you are doing it??

Answer 4

*find..

Answer 5

\[\frac{ dy }{ dx }[3x^2y+\frac{ dy }{ dx }x^3]+[xy^3+3y^2\frac{ dy }{ dx }3x^2] = 1 +\frac{ dy }{ dx }\]

Answer 6

that's how i did it, i'm not sure if i made an algebra mistake

Answer 7

Why are you using dy/dx with the first term??

Answer 8

what are you referring to?
I use dy dx because it said to find the derivative of y in respect to x

Answer 9

Let us take the first term first..

Do it slowly..

So we have : \(x^3y\).. Find its derivative..

Answer 10

\[x^3 y = x^3 \cdot \frac{dy}{dx} + 3x^2 \cdot y\]

Isn't it??

Answer 11

well it's basically what i wrote early except i switched the order of the number i took the derivative of.

Answer 12

earlier*

Answer 13

But you have attached dy/dx with that term..

Okay I understand your feelings that you are just showing that is the derivative only..

Answer 14

But in actual attaching dy/dx with that term will make your question wrong..

Answer 15

Well since i'm not sure how to solve it, would you be able to lay out the steps in which it is correctly solved?

Answer 16

And derivative of second term will be:

\[3xy^3 = 3[x \cdot 3y^2 + y^3]\]

Answer 17

Just wait...

To arrange steps it will take time.. Just few minutes..

Answer 18

but what about the y prime? am i not suppose to include it?

Answer 19

yes I forgot there, y' will be there with y^2..

Answer 20

\[\frac{d}{dx}(x^3y) + \frac{d}{dx}(3xy^3) = \frac{d}{dx}(x + y)\]

So this we have to solve as our question is saying..

Answer 21

See, firstly you know fully about the product rule and how to apply it, then it will be very simple..

Answer 22

Internet is not working properly..

Answer 23

alright so what we'll have after the product rule is going to be \[[3x^2y+y'x^3]+3[y^3+3y^2y'x]=1+y'\]

Answer 24

Is this correct?

Answer 25

So let us go further..

\[\implies x^3y' + 3yx^2 + 3(x \cdot 3y^2 \cdot y' + y^3) = 1 + y'\]

Answer 26

Yeah that is right..

Answer 27

and then i subract the 3yx^2 to the right and then subtract y' from the left right?

Answer 28

and then i'm stuck after that

Answer 29

Sorry, my internet connection is somewhat slow..

Answer 30

Now proceeding further we can separate y' terms and normal terms like this :

\[3x^2y + 3y^3 - 1 = y' + x^3y' + 9xy^2y'\]

Answer 31

Getting this step??

I am just separating normal terms on left hand side and y' terms on right..

Answer 32

I am expecting a reply from your side @duh_itsme

Answer 33

oh yes i'm sorry
i do

Answer 34

and the i factor out the y' right?

Answer 35

Yep, so go ahead.. This step you will do..

Answer 36

wait, for the multiplication rule of the second part, shouldn't it be 27xy^2y'? since you have to distribute in the 3, 3x, and 3y^2?

Answer 37

oh never mind, just ignore what i just said.

Answer 38

Ha ha ha..

Are you okay??

Can you go further??

Answer 39

Just factor out y' from RHS and show me where we have arrived..

Answer 40

oh yeah, hahahah i figured it out. My brain just went wacko for a moment. lol!

Answer 41

y'(-1+x^3+9xy^2) is what i got after factoring out y'

Answer 42

I just made a mistake and I think you are right..

Answer 43

now...i'm confused......

Answer 44

Are you doing it your way or according to my equation??

Answer 45

your way

Answer 46

Okay then there are two minus mistake..

Answer 47

\[3x^2y + 3y^3 - 1 = y' -x^3y' - 9xy^2y'\]

Answer 48

This should be the equation ..

Have you got how??

Answer 49

Bringing positive terms from LHS will make them Negative on RHS..

Getting??

Answer 50

I actually got something different.

Answer 51

Can you show me??

Answer 52

give me a moment please.

Answer 53

Take your time..

And I am really very sorry, I am just making mistakes while solving, that is not good for me as well as for the person like you whom I am teaching..

Answer 54

it's alright, you're still helping me. :P that's what counts

Answer 55

Now let us just concentrate on it and solve it out..

Can you show me what you get and take your time, I am in no hurry..

Answer 56

so what i got was y'x^3+9xy^2y'-y' = -3x^2y-3y^3+1

Answer 57

i guess it's the same as yours except the negative is distributed differently.

Answer 58

I was thinking the same. the terms I gathered on LHS, you are just gathering them on RHS, and it is no wrong..

So you are right dear..

Answer 59

Yep..

Now factor out y' from this..

Answer 60

yeah and then divide the equation on both sides right?

Answer 61

Yep..

Answer 62

thanks.

Answer 63

\[y'(x^3 + 9xy^2 - 1) = 1 - 3x^2y - 3y^3\]

Now just divide and find \(y'\)..

Answer 64

You are welcome dear..