Implicit Diff Example

[Jason76]

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{[y][1] - [x][y']}{[y]^{2}} = \dfrac{3y'}{3x - y}\) Next move - we need to isolate \(\displaystyle y'\)

 

[Jason76]

How about this?

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{[y][1] - [x][y']}{[y]^{2}} = \dfrac{3y'}{3x - y}\)

\(\displaystyle \dfrac{y - xy'}{[y]^{2}} = \dfrac{3y'}{3x - y}\)

\(\displaystyle y - xy' = \dfrac{3y'(y^{2})}{3x - y}\)

\(\displaystyle (3x - y)(y - xy') = 3y'(y^{2})\)

\(\displaystyle 3xy + 3x^{2}y' - y^{2} + xyy' = 3y'(y^{2})\) ??

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2} + xyy'}{y^{2}} = 3y'\)

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{y^{2}} = 3y' - xyy'\)

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{y^{2}} = y'[3 - xy]\)

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{(y^{2})(3 - xy)} = y'\)

Pushed to the other side as:

\(\displaystyle y' = \dfrac{3xy + 3x^{2}y' - y^{2}}{(y^{2})(3 - xy)}\) ??

\(\displaystyle y' = \dfrac{-6 - 8 - 1}{6 - 8}\)

\(\displaystyle y' = \dfrac{-15}{-2}\)

\(\displaystyle y' = \dfrac{15}{2}\)

\(\displaystyle y - 4 = \dfrac{15}{2}(x - 3)\)

\(\displaystyle y - 4 = \dfrac{15}{2}x -\dfrac{45}{2}\) ??

 

[HallsofIvy]

How about this?

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{[y][1] - [x][y']}{[y]^{2}} = \dfrac{3y'}{3x - y}\)

\(\displaystyle \dfrac{y - xy'}{[y]^{2}} = \dfrac{3y'}{3x - y}\)

Okay, this is just what you had before.

\(\displaystyle y - xy' = \dfrac{3y'(y^{2})}{3x - y}\)

\(\displaystyle (3x - y)(y - xy') = 3y'(y^{2})\)

And here you have multiplied by the two denominators to eliminate the fractions.

\(\displaystyle 3xy + 3x^{2}y' - y^{2} + xyy' = 3y'(y^{2})\) ??

Yes, this correct. On the left you have \(\displaystyle (3x- y)(y- xy')= 3x(y- xy')- y(y- xy')= 3xy- 3x^2y'- y^2+ xyy'\)

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2} + xyy'}{y^{2}} = 3y'\)

I would NOT do this- just because I don't like fractions! Instead get all "y'" terms on the same side, all that do not involve y' on the other.

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{y^{2}} = 3y' - xyy'\)

And, see, here that fraction has caused you to make an error. You subtracted "xyy'" on the right but subtracted the fraction \(\displaystyle \frac{xyy'}{y^2}\) on the left. You cannot just subtract part of the numerator of a fraction.
Of course, the rest of this is now wrong.

Instead, going back to \(\displaystyle 3xy + 3x^{2}y' - y^{2} + xyy' =3y'(y^{2})\), subtract \(\displaystyle 3x^2y'+ xyy'\) from both sides: \(\displaystyle 3xy- y^2= 3y^2y'- 3x^2y'- xyy'= (3y^2- 3x^2- xy)y'\)
and then, of course, divide both sides by \(\displaystyle 2y^2- 3x^2- xy\).

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{y^{2}} = y'[3 - xy]\)

\(\displaystyle \dfrac{3xy + 3x^{2}y' - y^{2}}{(y^{2})(3 - xy)} = y'\)

Pushed to the other side as:

\(\displaystyle y' = \dfrac{3xy + 3x^{2}y' - y^{2}}{(y^{2})(3 - xy)}\) ??

\(\displaystyle y' = \dfrac{-6 - 8 - 1}{6 - 8}\)

\(\displaystyle y' = \dfrac{-15}{-2}\)

\(\displaystyle y' = \dfrac{15}{2}\)

\(\displaystyle y - 4 = \dfrac{15}{2}(x - 3)\)

\(\displaystyle y - 4 = \dfrac{15}{2}x -\dfrac{45}{2}\) ??

 

[srmichael]

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{[y][1] - [x][y']}{[y]^{2}} = \dfrac{3y'}{3x - y}\) Next move - we need to isolate \(\displaystyle y'\) WRONG!

More specifically, the derivative of ln(3x - y) is wrong.

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{[y][1] - [x][y']}{[y]^{2}} = \dfrac{3 - y'}{3x - y}\)

 

[Jason76]

Let's give this another shot:

Using implicit differentiation and solving with respect to \(\displaystyle x\):

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{1}{3x - y}(x - y')\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{x - y'}{3x - y}\) What move now? We want to get the two \(\displaystyle y'\) terms to together, so we can factor them out. After that, we would try to isolate \(\displaystyle y'\)

 

[Deleted member 4993]

Let's give this another shot:

Using implicit differentiation and solving with respect to \(\displaystyle x\):

\(\displaystyle e^{x/y} = 3x - y\) ..... I'll do the following

\(\displaystyle \displaystyle e^{\frac{x}{y}} * \frac{y - x*y'}{y^2} = 3 - y'\)

\(\displaystyle \displaystyle e^{\frac{x}{y}} * \frac{1}{y} - \frac{x*y'}{y^2} * e^{\frac{x}{y}} \ = \ 3 - y'\)

\(\displaystyle \displaystyle 3 - e^{\frac{x}{y}} * \frac{1}{y} \ = \frac{x*y'}{y^2} * e^{\frac{x}{y}} \ - y'\)

\(\displaystyle \displaystyle 3 - e^{\frac{x}{y}} * \frac{1}{y} \ = \ y' * \left [\frac{x}{y^2} * e^{\frac{x}{y}} \ - 1 \right ]\)

\(\displaystyle \displaystyle 3y^2 - e^{\frac{x}{y}} * y \ = \ y' * \left [x * e^{\frac{x}{y}} \ - y^2 \right ]\)

Not any simpler - just different

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{1}{3x - y}(x - y')\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{x - y'}{3x - y}\) What move now? We want to get the two \(\displaystyle y'\) terms to together, so we can factor them out. After that, we would try to isolate \(\displaystyle y'\)

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[Deleted member 4993]

Let's give this another shot:

Using implicit differentiation and solving with respect to \(\displaystyle x\):

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{1}{3x - y}(x - y')\) ................Incorrect ............ \(\displaystyle \dfrac{d}{dx}\left [\dfrac{x}{y}\right ] \ne \dfrac{1}{y'}\)

\(\displaystyle \dfrac{1}{y'} = \dfrac{x - y'}{3x - y}\) What move now? We want to get the two \(\displaystyle y'\) terms to together, so we can factor them out. After that, we would try to isolate \(\displaystyle y'\)

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[Jason76]

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The answer you got in red (when you tried to solve without taking the \(\displaystyle \ln\) of both sides) didn't come out right on online homework. Or does it need to be simplified more?