How to differentiate the expression 4xy = y^2 + 2ln(x)

[footprints]

$ 4xy = y^2 + 2 \ln x
How do I differentiate that?

 

[ToxicBug]

Use implicit differentiation.

4y + 4xy' = 2yy' + 2/x
y'(4x - 2y) = 2/x - 4y
y' = (2/x - 4y)/(4x - 2y)

 

[footprints]

Use implicit differentiation.

4y + 4xy' = 2yy' + 2/x
y'(4x - 2y) = 2/x - 4y
y' = (2/x - 4y)/(4x - 2y)

Sorry, but I've not learned implicit differentiation. Haven't even heard of it.

 

[ToxicBug]

Then I don't know how you would isolate y in that to solve it the normal way.

 

[t!m]

Rewrite the expression as y^2 - 4xy + 2 \ln x = 0 and use the quadratic formula to find y in terms of x. Then derive.

 

[footprints]

Got it. Thank you.

 

[ToxicBug]

Rewrite the expression as y^2 - 4xy + 2 \ln x = 0 and use the quadratic formula to find y in terms of x. Then derive.

Why do teachers ask their students to do something like that when it can be solved using a more proficient technique?

 

[Hurkyl]

It's always reassuring to see that two drastically different methods yield the same results.

Also, doing this problem both ways will allow students to greater appreciate implicit differentiation, once they see how much work it can save.