How Do I Differentiate the Term 6xy in an Implicit Differentiation?

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To differentiate the term 6xy in the implicit differentiation of the equation x^3 + y^3 = 6xy, the product rule is applied, resulting in 6y + 6xy'. The user initially struggles with this differentiation but realizes the correct approach involves treating 6x as one function and y as another. After applying the product rule, they arrive at the equation x^2 + y^2y' = 2y + 2xy'. The discussion concludes with the user acknowledging their understanding and deciding to take a break.
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Homework Statement



Hey all!

Ok, doing a simple implicit differential.


x^3 + y^3 = 6xy

The 6xy is messing me up! How do I differentiate that??

The book says 6xy turns into 6y + 6xy'

I do not understand how to differentiate 6xy (how the x and y are stuck together)

Ok, doing a simple implicit differential.


Homework Equations





The Attempt at a Solution

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

Notice how it's the product rule?
 
Yes I know its the product rule...

Wait. Ahh, I see!

6x is f and y is g (in terms of f'g + fg')

which gives me

(6y) + 6x(y')

:)

Thanks!
 
But now that brings me to my 2nd question.

This leaves me with: x^2 + y^2y' = 2y + 2xy'

I don't understand how to solve for y' when I have 2 of them.
 
Use algebra! Move both y' terms to one side of the equation everything else to the other side and solve for it.
 
ahhg, I think I need to stop. I am forgetting such silly things!

I got the question. Thanks guys. Think its time for little break :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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