Why Does My Derivative Calculation for f(x)=x^3ln(x^2) Differ from the Textbook?

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Homework Help Overview

The discussion revolves around the differentiation of the function f(x) = x^3 ln(x^2). Participants are examining the application of the product rule in calculus and comparing their results with a textbook answer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the product rule but questions their result compared to the textbook. Other participants discuss the derivative of ln(x) and its properties, raising questions about the differentiation of logarithmic functions.

Discussion Status

The discussion is active, with participants exploring different aspects of logarithmic differentiation. Some guidance on properties of logarithms has been offered, but there is no explicit consensus on the original poster's confusion regarding their derivative calculation.

Contextual Notes

Participants are navigating through potential misunderstandings related to the properties of logarithms and their derivatives, particularly in the context of the original function's differentiation.

Ry122
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f(x)=x^3lnx^2
product rule
y'=u'v+v'u
u=x^3
u'=3x^2
v=lnx^2
v'=2lnx^2

f'(x)=3x^2(lnx^2)+2x^3lnx^2)

but answer in book is f'(x)=3x^2(lnx^2)+2x^2
What am i doing wrong?
 
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What's the derviative of just ln(x)?
 
lnx^b=blnx
x has an exponent of 1
 
d/dx lnx = 1/x
 
So does the derivative of lnx^b not = blnx?
 
Ry122 said:
So does the derivative of lnx^b not = blnx?

ln(x^b) = b * ln(x)

This is a property of logarithms and has nothing to do with differentiation.

d/dx [b * ln(x)] = b/x
 
You can also find the derivative if you use a u-sub rather than simplifying the natural log. You might notice that the 1 in the numerator is the derivative of the denominator.

Try [tex]\frac{u'}{u}[/tex]
 

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