Question

I need help with this question Please Urgently!! Find dy/dx: x^2-xy+y^3=1.

Answer 1

Please anyoneeee I really need a help

Answer 2

okay so you will take the derivative of each thing like...

2x-y+x(dy/dx)+3y^2(dy/dx)=0
get the dy.dx on one side

Answer 3

x(dy/dx)+3y^2(dy/dx)=y-2x
factor dy/dx

Answer 4

dy/dx(x+3y^2)=y-2x

Answer 5

dy/dx= (y-2x)/(x+3y^2)

Answer 6

Good c:
I see one minor boo boo though,\[\huge \frac{d}{dx}(-xy)=-y\color{orangered}{-}x\frac{dy}{dx}\]
@cherio12

Answer 7

ohh where is it?

Answer 8

ya good call

Answer 9

forgot to distribute the negative

Answer 10

but where I cant see that which one?

Answer 11

The fancy math not showing up for you pinky? :O

The derivative of the second term, the -xy. It should produce 2 negative terms.

Answer 12

the denominator should be 3y^2-x

Answer 13

opps lolz sorry thanks i am blind :P

Answer 14

Sorry one more question if anyone can help pleaseeeeee

Answer 15

\[xe ^{2y}=5x+y ^{2}\]

Answer 16

you want the derivative,correct?

Answer 17

the 1st degree partials

Answer 18

No not partials :) This is Calc 1. Implicit Differentiation.

Answer 19

or is this a differential equation?

Answer 20

yes I want derivative its an implicit differentiation

Answer 21

Taking the derivative of the left side, here is the setup for product rule,\[\large \left(xe^{2y}\right)'=(x)'e^{2y}+x(e^{2y})'\]

Answer 22

hmm ok

Answer 23

Really? Already confused? :(

Hmm you should have learned product rule at this point <:o hmmm

Answer 24

^^^ lol take it easy

Answer 25

no no not confuse lolz sorry

Answer 26

I m following you

Answer 27

What sweet...?

Answer 28

differentiate both sides with respect to x. The left side requires the product rule, and so do the two terms on the right side.

Answer 29

Here's the left side\[\frac{ \delta }{ \delta x }xe^{2y} = e^{2y}+2x(\frac{ dy }{ dx })e^{2y}\]

Answer 30

ok

Answer 31

The process is the same for the right hand side. As you can see, we had to implicitly differentiate with respect to x, since we couldn't explicitly differentiate with respect to x, since y, being a function of x, is unknown.

Answer 32

ok

Answer 33

I understand, I skipped pre-calc and trig and went straight to calc 1. I didn't fully understand most of the material until calc 2 lol

Answer 34

ohh :(

Answer 35

I m really upset thought its due tomorrow :(

Answer 36

hey hey, you just have to clear your mind, relax, and look at the material anew, ignoring all frustration and anxiety.

Answer 37

Do you want me to explain what implicit differentiation is?

Answer 38

yes please and if you can also help me to solve this whole problem I will appreciate your help :(

Answer 39

well the answer to this problem is:\[\frac{ \delta }{ \delta x}xe^{2y} = \frac{ \delta }{ \delta x}(5x +y^2)=\frac{ \delta }{ \delta x}(5x) + \frac{ \delta }{ \delta x}(y^2)\]\[e^{2y}+2x \frac{ \delta y }{ \delta x } e^{2y} = 5 + 2y(\frac{ \delta y }{ \delta x })\] isolate dy/dx:\[2x \frac{ \delta y }{ \delta x } e^{2y} - 2y(\frac{ \delta y }{ \delta x }) = 5 - e^{2y}\]\[\frac{ \delta y }{ \delta x } (2xe^{2y} - 2y) = 5 - e^{2y}\]\[\frac{ \delta y }{ \delta x } = \frac{ 5 - e^{2y} }{ (2xe^{2y} - 2y) }\]

Answer 40

seems complicated, but it really isn't. Math notation makes everything more difficult. I'm using delta x instead of dx because y might be a function of other variables besides x

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