Calculating Gradient of ln(r) in 3D Space

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SUMMARY

The discussion focuses on calculating the gradient of the function f=ln(r) in 3D space, where r=sqrt(x^2+y^2+z^2). The correct application of the chain rule reveals that the gradient with respect to x is indeed df/dx = x/r^2, not x/r. The additional factor of 1/r arises from the differentiation of the logarithmic function and the proper application of the chain rule. Participants clarify the steps involved in deriving the gradient, emphasizing the importance of careful differentiation.

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  • Understanding of multivariable calculus, specifically gradient calculations.
  • Familiarity with the chain rule in differentiation.
  • Knowledge of logarithmic functions and their properties.
  • Basic comprehension of 3D coordinate systems and distance formulas.
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  • Study the application of the chain rule in multivariable calculus.
  • Learn about gradient vectors and their significance in vector calculus.
  • Explore the properties of logarithmic differentiation in various contexts.
  • Investigate the implications of gradients in physical applications, such as optimization problems.
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Students and professionals in mathematics, physics, and engineering who are working with multivariable functions and require a solid understanding of gradient calculations in 3D space.

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So, the question is: find the gradient of f=ln(r), where r=sqrt(x^2+y^2+z^2)^(1/2).

For the partial with respect to x, I use the chain rule: df/dr*dr/dx.

df/dr=1/r
dr/dx=(1/2)*(2x)=x

Which would give df/dx = x/r.

But the book gets x/r^2

Where does the extra 1/r come from?
 
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brentd49 said:
So, the question is: find the gradient of f=ln(r), where r=sqrt(x^2+y^2+z^2)^(1/2).
...
dr/dx=(1/2)*(2x)=x
...
This isn't right. Remember to use the chain rule here as well.
 
hah. of course. thanks
 

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