Implicitly differentiating PDE (multivariable calculus)

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To find dz/dx at the point (1,1,1) for the equation xy + z^3x - 2yz = 0, implicit differentiation is necessary. The correct approach involves applying the product rule and the chain rule, particularly for the term z^3x, which results in 3z^2x(dz/dx) + z^3. A miscalculation was initially made regarding the differentiation process, leading to confusion about the final equation. The importance of correctly applying the chain rule was emphasized, clarifying the differentiation steps. Understanding these rules is crucial for accurately solving multivariable calculus problems.
Legion81
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The problem:
Find the value of dz/dx at the point (1,1,1) if the equation xy+z3x-2yz=0 defines z as a function of the two independent variables x and y and the partial derivative exists.

I don't know how to approach the z3x part. I thought you would use the product rule and get 3(dz/dx)2x + z3. But if that is right, the final equation looks something like

y + 3x(dz/dx)2 + z3 - 2y(dz/dx) = 0

And I don't think that is right. The only way I know to solve that would be with the quadratic equation and that gives a complex value. Am I forgeting the chain rule somewhere or just way off on approaching this problem?
 
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I have done very little multivariable so this could easily be wrong, but if you implicitly differentiated z^3x shouldn't you get 3z^2x*dz/dx + z^3?
 
Frillth said:
I have done very little multivariable so this could easily be wrong, but if you implicitly differentiated z^3x shouldn't you get 3z^2x*dz/dx + z^3?

That's right because you have the 3z2x(dz/dx) + z3(dx/dx). I don't know what I was thinking... Thank You!
 
You didn't apply the chain rule, remember df(u)/dx = f(u)'*du/dx
 
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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