# Derivatives of ln functions

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?

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 $$\frac{u'}{u}$$