How do you find the derivative of #ln(x^3+3x)#?

1 Answer
Steve M
Oct 17, 2016

# d/dxln(x^3+3x)=(3x^2+3)/(x^3+3x) #

Explanation:

We us the chain rule: # dy/dx=dy/(du)(du)/dx #

Let #y=ln(x^3+3x)#, and
Let #u=x^3+3x#

# y=ln(x^3+3x) #
# :. y=lnu #
# :. dy/(du)=1/u #

# u=x^3+3x #
# :. (du)/dx=3x^2+3 #

So using the chain rule we have:
# dy/dx=dy/(du)(du)/dx #
# :. dy/dx=1/u(3x^2+3) #
# :. dy/dx=(3x^2+3)/(x^3+3x) #

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