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OpenStudy (anonymous):

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

OpenStudy (anonymous):

Please anyoneeee I really need a help

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

zepdrix (zepdrix):

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

OpenStudy (anonymous):

ohh where is it?

OpenStudy (anonymous):

ya good call

OpenStudy (anonymous):

forgot to distribute the negative

OpenStudy (anonymous):

but where I cant see that which one?

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.

OpenStudy (anonymous):

the denominator should be 3y^2-x

OpenStudy (anonymous):

opps lolz sorry thanks i am blind :P

OpenStudy (anonymous):

Sorry one more question if anyone can help pleaseeeeee

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you want the derivative,correct?

OpenStudy (anonymous):

the 1st degree partials

zepdrix (zepdrix):

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

OpenStudy (anonymous):

or is this a differential equation?

OpenStudy (anonymous):

yes I want derivative its an implicit differentiation

zepdrix (zepdrix):

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

OpenStudy (anonymous):

hmm ok

zepdrix (zepdrix):

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

OpenStudy (anonymous):

^^^ lol take it easy

OpenStudy (anonymous):

no no not confuse lolz sorry

OpenStudy (anonymous):

I m following you

zepdrix (zepdrix):

What sweet...?

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok

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.

OpenStudy (anonymous):

ok

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

OpenStudy (anonymous):

ohh :(

OpenStudy (anonymous):

I m really upset thought its due tomorrow :(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Do you want me to explain what implicit differentiation is?

OpenStudy (anonymous):

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

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

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

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