Not understanding last steps in an implicit differentiation problem.

[TheVee]

Hi all,
I'm trying to keep my brain functioning during these extremely trying times. It's been over 50 years since I had calculus in college. But I've always been fascinated with math and science.
So I decided to delve back into calculus. One of the topics in our text book back then consisted of one paragraph about implicit differentiation. However, the prof spent about a week on the subject. I never really got a good handle on it and wanted to see if I could understand it better and try my hand at solving some problems.
I downloaded some problems from the 'Net and tried solving. I ran into one problem that I tried. I came up with an answer, but decided to use one of the online calculators which also shows the steps to solving. You can find that calculator at https://www.symbolab.com/solver/implicit-derivative-calculator.

The problem is: sin(x/y)=1/2

I was able to understand the steps up to:

cos(x/y)(-xy' +y) = 0

y2

The next step is shown as: cos(x/y)(-xy' +y) = 0 citing the Zero Factor Principle: If ab=0 then a=0 or b=0 (or both a=0 and b=0)
I'm just not sure what happened to the y². Did they multiply both sides by y²?

Once again, for the following step, they cite the Zero Factor Principle. But only show: -xy' + y = 0. So did they assume cos(x/y) was zero? Or did they divide each side by it?

From there they finalize the solution as y' = y/x.

Can anyone clarify the two questions I posed above?

Thanks,
Vic

 

[Harry_the_cat]

1.

For a fraction to be 0, the numerator has to be zero as long as the denominator isn't. \(\displaystyle y^2\) is not 0, since y can't be, otherwise the original problem would not be defined. The other way of looking at it is to multiply both sides by \(\displaystyle y^2\).

 

[Harry_the_cat]

2.

If cos(x/y) =0 then sin(x/y)=1 or -1, BUT sin(x/y) = 1/2 (from original statement) so cos(x/y) can't be 0.
This means that the second factor (y-xy')=0

 

[pka]

1.
View attachment 18072

Using LaTeX here is standard to use prime. y^{\prime} gives \(y^{\prime}\)

 

[TheVee]

Harry_the_cat, thanks for your replies and explanations. Your first response confirmed my suspicion. Your second response was extremely helpful and hopefully expanded my awareness of what I need to consider going forward. I had never looked at the fact that the sin(x/y) is not 0, therefore cos(x/y) is not 0. DOH!!

pka, thanks for your response. I'm not familiar with LaTeX. I'll try looking into it.

Thanks to you both!
Vic