Partial derivatives of a function

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SUMMARY

The discussion focuses on finding the first-order partial derivatives of the function ln((√(x²+y²) - x)/(√(x²+y²) + x)). The user successfully separates the logarithm into a subtraction of two logarithmic terms. They apply the derivative rule for logarithms, specifically d ln(u) = 1/u, and are guided to take the partial derivatives with respect to x and y. The key takeaway is the application of the chain rule in conjunction with logarithmic differentiation to simplify the process of finding partial derivatives.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with logarithmic differentiation
  • Knowledge of the chain rule in calculus
  • Basic proficiency in handling square roots and algebraic manipulation
NEXT STEPS
  • Practice finding partial derivatives of more complex functions
  • Explore the application of the chain rule in multivariable calculus
  • Study the properties of logarithmic functions in calculus
  • Learn about gradient vectors and their significance in optimization problems
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable functions and partial derivatives, as well as educators looking for examples of logarithmic differentiation techniques.

Mr.Rockwater
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1. The problem statement, all variables and given known data

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

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

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 [itex]\sqrt{x²+y²}[/itex] . 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?
 
Last edited:
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You were on the right track.
[itex]ln((\sqrt{(x^2+y^2} - x))-ln((\sqrt{x^2+y^2} + x))[/itex]
You can now take the partial derivative of this function with respect to x, then respect to y.

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

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