How do you differentiate $ln[ (2x^3)-(3x^2)+(7) ]$ ?

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Answer 1 · 2018-07-06T09:05:48

$1/(2*x^3-3*x^2+7)*(6*x^2-6*x)$

Using the fact that $(ln(x))'=1/x$ and the chain rule we get

$1/(2x^3-3x^2+7)*(6x^2-6x)$

How do you differentiate ln[ (2x^3)-(3x^2)+(7) ] ln[ (2x^3)-(3x^2)+(7) ]?