How Do You Differentiate Functions Involving Natural Logarithms?

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SUMMARY

The discussion focuses on differentiating the function \(\frac{V_0}{x \ln(\frac{b}{x})}\) with respect to \(x\). The user initially struggled with the differentiation of the natural logarithm and utilized MATLAB for assistance. The correct differentiation yields the expression \(\frac{-V_0}{x^2 \ln(b/x)} + \frac{V_0 \cdot b}{x^3 \ln(b/x)^2}\), although the user questioned the presence of \(b\) in the second term. The discussion also references implicit differentiation techniques for logarithmic functions.

PREREQUISITES
  • Understanding of calculus, specifically differentiation techniques.
  • Familiarity with natural logarithms and their properties.
  • Basic knowledge of MATLAB for computational verification.
  • Experience with implicit differentiation methods.
NEXT STEPS
  • Review the rules of differentiation for logarithmic functions.
  • Practice implicit differentiation with various logarithmic equations.
  • Explore MATLAB's symbolic toolbox for differentiation tasks.
  • Study the applications of logarithmic differentiation in real-world problems.
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Students studying calculus, educators teaching differentiation techniques, and anyone seeking to improve their understanding of logarithmic functions and their derivatives.

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[SOLVED] need some help differentiating

Uh my bad, forgot how to differentiate ln's


Homework Statement


[tex]\frac{V_0}{xln(\frac{b}{x})}[/tex]
Find dE/dx

I can almost get the answer, but I had to use MATLAB to find the actual answer, so I am kind of feeling stupid now.

My problem is when I try to differentiate the ln

Homework Equations





The Attempt at a Solution



[tex]d/dx(\frac{V_0}{xln(\frac{b}{x})}) = \frac{-V_0}{x^2 ln(b/x)} + \frac{V_0 * b/x^2}{x ln(b/x)^2} =[/tex]

[tex]\frac{-V_0}{x^2 ln(b/x)} + \frac{V_0 * b}{x^3 ln(b/x)^2}[/tex]

The second term should only have x^2 in it (and no b), does anyone see where I went wrong?
 
Last edited:
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[tex]log_a(x)=y=>a^{y}=x[/tex] let's differentiate this implicitly, we get:

[tex]\frac{d}{dx}(a^{y}=x)=>\frac{dy}{dx}a^{y}lna=1=>\frac{dy}{dx}=\frac{1}{a^{y}lna}=>\frac{dy}{dx}=\frac{1}{xlna}[/tex]

When a=e, we get:[tex]\frac{dy}{dx}=\frac{1}{x}[/tex]. So it is just a special case of a general case.
 

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