Partial derivatives of a function

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To find the first-order partial derivatives of the function ln((√(x²+y²) - x)/(√(x²+y²) + x)), the logarithm can be separated into two terms for easier differentiation. The derivative rule for logarithms, d ln(u) = 1/u, is applied, leading to a more complex expression involving x², y², and √(x²+y²). The discussion emphasizes taking the partial derivatives with respect to both x and y after simplifying the logarithmic expression. The key takeaway is to utilize the quotient rule and the chain rule effectively in the differentiation process.
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1. The problem statement, all variables and given known data

Find the partial derivatives (1st order) of this function:

ln((\sqrt{(x^2+y^2} - x)/(\sqrt{x^2+y^2} + x))

Homework Equations





The Attempt at a Solution



I obviously separated the logarithm quotient into a subtraction, then applied the rule d ln(u) = 1/u. However, what I end up with is four terms with a bunch of x²+y² and \sqrt{x²+y²} . I'm just starting out with partial derivatives so is there any obvious trick that I'm not familiar with in this type of situation?
 
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You were on the right track.
ln((\sqrt{(x^2+y^2} - x))-ln((\sqrt{x^2+y^2} + x))
You can now take the partial derivative of this function with respect to x, then respect to y.

remember:
\frac{\partial ln[f(x, y)]}{\partial x}=\frac{1}{f(x, y)}\frac{\partial f(x, y)}{\partial x}
 
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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