OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

Please anyoneeee I really need a help

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

ohh where is it?

4 years ago
OpenStudy (anonymous):

ya good call

4 years ago
OpenStudy (anonymous):

forgot to distribute the negative

4 years ago
OpenStudy (anonymous):

but where I cant see that which one?

4 years ago
zepdrix (zepdrix):

The fancy math not showing up for you pinky? :O The derivative of the second term, the -xy. It should produce 2 negative terms.

4 years ago
OpenStudy (anonymous):

the denominator should be 3y^2-x

4 years ago
OpenStudy (anonymous):

opps lolz sorry thanks i am blind :P

4 years ago
OpenStudy (anonymous):

Sorry one more question if anyone can help pleaseeeeee

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

you want the derivative,correct?

4 years ago
OpenStudy (anonymous):

the 1st degree partials

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

or is this a differential equation?

4 years ago
OpenStudy (anonymous):

yes I want derivative its an implicit differentiation

4 years ago
zepdrix (zepdrix):

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

4 years ago
OpenStudy (anonymous):

hmm ok

4 years ago
zepdrix (zepdrix):

Really? Already confused? :( Hmm you should have learned product rule at this point <:o hmmm

4 years ago
OpenStudy (anonymous):

^^^ lol take it easy

4 years ago
OpenStudy (anonymous):

no no not confuse lolz sorry

4 years ago
OpenStudy (anonymous):

I m following you

4 years ago
zepdrix (zepdrix):

What sweet...?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

ok

4 years ago
OpenStudy (anonymous):

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.

4 years ago
OpenStudy (anonymous):

ok

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

ohh :(

4 years ago
OpenStudy (anonymous):

I m really upset thought its due tomorrow :(

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

Do you want me to explain what implicit differentiation is?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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:\[2x \frac{ \delta y }{ \delta x } e^{2y} - 2y(\frac{ \delta y }{ \delta x }) = 5 - e^{2y}\]\[\frac{ \delta y }{ \delta x } (2xe^{2y} - 2y) = 5 - e^{2y}\]\[\frac{ \delta y }{ \delta x } = \frac{ 5 - e^{2y} }{ (2xe^{2y} - 2y) }\]

4 years ago
OpenStudy (anonymous):

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

4 years ago
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