Implicit Differentiation Clarification

[Asphyxiated]

I am hoping my addition of binomials is correct:

xy^{2} - 3x^{2}y +x =1

\frac {d(xy^{2}-3x^{2}y+x)}{dx} = \frac {d(1)}{dx}

(2yx\frac{dy}{dx}+y^{2}) - (3x^{2}\frac{dy}{dx}+6xy)+1 = 0

then I change the subtraction to addition and move the 1:

(2yx\frac{dy}{dx}+y^{2}) + (-3x^{2}\frac{dy}{dx}-6xy) = -1

2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} +y^{2}-6xy = -1

2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} +y^{2} = -1 + 6xy

2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} = -1 + 6xy - y^{2}

\frac{dy}{dx}(2yx - 3x^{2}) = -1 + 6xy - y^{2}

\frac{dy}{dx}= \frac {-1 + 6xy - y^{2}} {2yx - 3x^{2}}

 

[Dick]

I'm with you up to here:
<br /> \frac{dy}{dx}= \frac {-1 + 6xy - y^{2}} {2yx - 3x^{2}} <br />

That's good. It's actually good work, then you went completely off the radar using some algebra rule you apparently just made up. Will you stop doing that?
This is getting to the point were I'm going to have to ask you not only to show your steps but to show the justification for each step. Otherwise, you are never going to learn to self-check these steps. You do fine for a while and then you just make a completely bongo move. Do you know that the manipulations you do in algebra are largely confined to i) adding or subtracting the same quantity from both sides an equation and ii) multiplying or dividing both sides of an equation by the same quantity? They aren't just like, randomly rearranging symbols in vaguely plausible ways.

How would you justify the step after that where you just bleeped out an xy from the numerator and denominator?

 

[Asphyxiated]

well I checked everything up to the point that I messed up, I thought that same values in the numerator and the denominator cancel. Since x*y = y*x and x=x and y=y. I definitely didn't just rearrange the symbols in vaguely plausible ways... but after checking the simplification (i will call it that) i see that I was wrong in that regard. Really the problem lies in me having to recall the correct way some step is done that I haven't used much recently, and vaguely remember from high school, I removed the two last steps since they are wrong

Not that it matters much since I evidently made up my own rule on this problem but did you look at the problem posted before this one? Pretty sure that one is 100% the first time around. I should just stop making assumptions when posting a problem. The last result I have on my board for the problem you just posted about is where you said that it was good. Usually I'll think that I see something else that can be done after I get it all written up and don't bother to check at that point... So, yeah

 

[Dick]

Sure. The main point is you can add/subtract to both sides of equation as in a=b+c -> a-c=(b+c)-c -> a-c=b or you can multiply/divide in a similar pattern. That's all. Ask yourself if that's REALLY what you are doing before you do anything. Cancelling the xy doesn't fall into either of those categories. Then maybe you won't have to check so much.

 

[Asphyxiated]

2xy^{-2}+x^{-2} =y

-4xy^{-3}\frac{dy}{dx} + 2y^{-2} - 2x^{-3} = \frac {dy}{dx}

-4xy^{-3}\frac{dy}{dx} + 2y^{-2} = \frac {dy}{dx} + 2x^{-3}

-4xy^{-3}\frac{dy}{dx} = \frac {dy}{dx} + 2x^{-3}- 2y^{-2}

-4xy^{-3}\frac{dy}{dx} -\frac {dy}{dx} = 2x^{-3}- 2y^{-2}

\frac{dy}{dx}(-4xy^{-3} -1) = 2x^{-3}- 2y^{-2}

\frac{dy}{dx} = \frac {2x^{-3}- 2y^{-2}} {-4xy^{-3} -1}

 

[Dick]

Looks ok to me.

 

[Asphyxiated]

y = sin(xy)

\frac {dy}{dx} = cos(xy) * (x\frac{dy}{dx} +y)

\frac {dy}{dx} = (x\frac{dy}{dx})cos(xy)+(y)cos(xy)

\frac {dy}{dx}(1 -(x)cos(xy))= (y)cos(xy)

\frac {dy}{dx}= \frac{y*cos(xy)} {1 -x*cos(xy)}

 

[Asphyxiated]

And another:

x = cos^{2} (y)

1 = (2cos(y))(-sin(y))(\frac{dy}{dx})

\frac {1} {(2cos(y))(-sin(y))} = \frac {dy}{dx}

 

[Asphyxiated]

Hopefully getting the hang of this:

y = e^{x+2y}

\frac {dy}{dx} = e^{x+2y} * (1+2\frac{dy}{dx})

\frac {dy}{dx} = e^{x+2y} + 2\frac{dy}{dx}e^{x+2y}

\frac {dy}{dx} - 2\frac{dy}{dx}e^{x+2y} = e^{x+2y}

\frac {dy}{dx}(1 - 2e^{x+2y}) = e^{x+2y}

\frac {dy}{dx} = \frac {e^{x+2y}} {(1 - 2e^{x+2y})}

 

[Dick]

Yes, I think you've gotten the hang of it.

 

[Asphyxiated]

One more for good measure? It's the last question in the section any way

y^{2} = ln (2x+3y)

2y\frac{dy}{dx} = \frac {1} {2x+3y} * (2+3\frac{dy}{dx})

2y\frac{dy}{dx} = \frac {2} {2x+3y} + \frac {3}{2x+3y}\frac{dy}{dx}

2y\frac{dy}{dx} - \frac {3}{2x+3y}\frac{dy}{dx} = \frac {2} {2x+3y}

\frac{dy}{dx}(2y - \frac {3}{2x+3y})= \frac {2} {2x+3y}

which is to say:

\frac{dy}{dx}(\frac {(2y)(2x+3y)}{2x+3y} - \frac {3}{2x+3y})= \frac {2} {2x+3y}

\frac{dy}{dx}(\frac {(2y)(2x+3y)-3}{2x+3y})= \frac {2} {2x+3y}

\frac{dy}{dx}= \frac {\frac {2} {2x+3y}} {\frac {(2y)(2x+3y)-3}{2x+3y}}

\frac{dy}{dx}= \frac {2} {2x+3y} *\frac {2x+3y}{(2y)(2x+3y)-3}

\frac{dy}{dx}= \frac {(2)(2x+3y)} {(2x+3y)(2y)(2x+3y)-3}

\frac{dy}{dx}= \frac{2} {(2y)(2x+3y)-3}

 

[Dick]

Last one?? Promise? That one looks good to me.

 

[Asphyxiated]

Yep that's the last one, thanks 1000x for all the help and patience Dick!

 

[Dick]

Very welcome. It's nice to see someone learn from mistakes and learn not to repeat them. Wish that happened more often around here. You are welcome back anytime!

 

[Asphyxiated]

Oh I am sure that I will have more questions soon enough as I am only on chapter 3 of 13 in this particular book, but I am glad to hear that I didn't ask too many questions as I had suspected.