Implicit differentiation question

[Bmanmcfly]

In the example I must complete, it turns out that I'm stuck with dy/dx on both sides of the equation, and includes a function using power rule, containing a dy/dx.

Now, what should I do? Factor out the dy/dx from the simpler side and have a complex sloppy looking equation, make the function equal 0 (since I cannot isolate the dy/dx on either side cleanly. I don't need to find a certain point.

Sorry for phrasing this like this, but I don't want the answer as much as how to get the answer, the one I keep getting can't seem to be factored easily to isolate the dy/dx and that feels incomplete.

Thanks for any help.

 

[srmichael]

In the example I must complete, it turns out that I'm stuck with dy/dx on both sides of the equation, and includes a function using power rule, containing a dy/dx.

Now, what should I do? Factor out the dy/dx from the simpler side and have a complex sloppy looking equation, make the function equal 0 (since I cannot isolate the dy/dx on either side cleanly. I don't need to find a certain point.

Sorry for phrasing this like this, but I don't want the answer as much as how to get the answer, the one I keep getting can't seem to be factored easily to isolate the dy/dx and that feels incomplete.

Thanks for any help.

They typical way of doing these is to bring all of the terms with the dy/dx on one side of the equal sign and then move all of the other terms without dy/dx to the other side. Then, factor out the dy/dx from the terms that have it and then solve for dy/dx by simply dividing by the factor that you have once you factored out the dy/dx.

For example, let's say you ended up with this after implicitly differentiating. Then solve for dy/dx like so:

\(\displaystyle 3x^2\frac{dy}{dx}+2xy = 4x^3-5xy^2\frac{dy}{dx}\)

\(\displaystyle 3x^2\frac{dy}{dx}+5xy^2\frac{dy}{dx}=4x^3-2xy\)

\(\displaystyle \frac{dy}{dx}(3x^2+5xy^2)=4x^3-2xy\)

\(\displaystyle \frac{dy}{dx}=\frac{4x^3-2xy}{3x^2+5xy^2}\)

\(\displaystyle \frac{dy}{dx}=\frac{x(4x^2-2y)}{x(3x+5y^2)}\)

\(\displaystyle \frac{dy}{dx}=\frac{4x^2-2y}{3x+5y^2}\)

 

[lookagain]

Then solve for dy/dx like so:

.
.
.

\(\displaystyle \frac{dy}{dx}=\frac{4x^3-2xy}{3x^2+5xy^2}\)

This isn't finished.


\(\displaystyle \frac{dy}{dx} = \frac{x(4x^2 - 2y)}{x(3x + 5y^2)}\)


\(\displaystyle \frac{dy}{dx} = \boxed{\frac{4x^2 - 2y}{3x + 5y^2}}\)

 

[Bmanmcfly]

They typical way of doing these is to bring all of the terms with the dy/dx on one side of the equal sign and then move all of the other terms without dy/dx to the other side. Then, factor out the dy/dx from the terms that have it and then solve for dy/dx by simply dividing by the factor that you have once you factored out the dy/dx.

For example, let's say you ended up with this after implicitly differentiating. Then solve for dy/dx like so:

\(\displaystyle 3x^2\frac{dy}{dx}+2xy = 4x^3-5xy^2\frac{dy}{dx}\)

\(\displaystyle 3x^2\frac{dy}{dx}+5xy^2\frac{dy}{dx}=4x^3-2xy\)

\(\displaystyle \frac{dy}{dx}(3x^2+5xy^2)=4x^3-2xy\)

\(\displaystyle \frac{dy}{dx}=\frac{4x^3-2xy}{3x^2+5xy^2}\)

\(\displaystyle \frac{dy}{dx}=\frac{x(4x^2-2y)}{x(3x+5y^2)}\)

\(\displaystyle \frac{dy}{dx}=\frac{4x^2-2y}{3x+5y^2}\)

Thanks, that worked.

Brings up a new question, is there a better situation (forgive I'm not sure how to get equations to show up proper), if I'm left with a situation like :

4(2y-x)^3(2dy/dx-1) ... =dy/dx that would not involve resolving the cube?

Reason being, the answer I got is to the effect of :
dy/dx= (32y^3-64y^2x-32yx^2-...2x)/(128y^3-64y^2x...-1)

I can't seem to figure out how to find factors of such a complex equation... Also I learned that I shouldnt have waited 10 years to get back into school, cause now I seem to be struggling to keep afloat.

Thanks again for the help...

 

[HallsofIvy]

Thanks, that worked.

Brings up a new question, is there a better situation (forgive I'm not sure how to get equations to show up proper), if I'm left with a situation like :

4(2y-x)^3(2dy/dx-1) ... =dy/dx that would not involve resolving the cube?

I'm not sure what the "..." indicates but for the rest, we have 8(2y- x)^3(dy/dx)- 4(2y- x)^3= dy/dx so that
(8(2y- x)^3- 1)dy/dx= 4(2y- x)^3

dy/dx= 4(2y- x)^3/(8(2y- x)^3- 1)

Reason being, the answer I got is to the effect of :
dy/dx= (32y^3-64y^2x-32yx^2-...2x)/(128y^3-64y^2x...-1)

I can't seem to figure out how to find factors of such a complex equation... Also I learned that I shouldnt have waited 10 years to get back into school, cause now I seem to be struggling to keep afloat.

Thanks again for the help...

 

[Bmanmcfly]

I'm not sure what the "..." indicates but for the rest, we have 8(2y- x)^3(dy/dx)- 4(2y- x)^3= dy/dx so that
(8(2y- x)^3- 1)dy/dx= 4(2y- x)^3

dy/dx= 4(2y- x)^3/(8(2y- x)^3- 1)

Sorry, the '...' was to shorten the actual equation (im trying to ask the question as general as possible because getting full answers won't be as helpful long term.)

Basically to show that the equation was quite complex.

I think I mistyped the first part... Or did I catch that right that I could just multiply the (dy/dx - 1) by the (2y-x)^3 directly and then factor it?

I didn't think that this would hold the same value? If it helps the "=dy/dx" at the end of the line was because of a dire stove of the (y+3) on the other side before differentiating, and not as a symbol for =answer.
That much was my bad, not only do I suck at math (from lack of practice), but I suck at math forums. Lol.