How do you find the derivative of ${ln (x^{3} + 3 x)}^{3}$ $ln(x^3+3x)^3$ ?

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How do you find the derivative of ${ln (x^{3} + 3 x)}^{3}$ $ln(x^3+3x)^3$ ?

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Answer 1 · 2015-04-07T19:41:54

I would use the Chain Rule after rearranging the $ln$ $ln$ as:
${ln (x^{3} + 3 x)}^{3} = 3 ln (x^{3} + 3 x)$ $ln(x^3+3x)^3=3ln(x^3+3x)$
So deriving you get:
$3 \cdot \frac{1}{x^{3} + 3 x} \cdot (3 x^{2} + 3) = 9 \frac{x^{2} + 1}{x (x^{2} + 3)}$ $3*1/(x^3+3x)*(3x^2+3)=9(x^2+1)/(x(x^2+3))$

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How do you find the derivative of ${ln (x^{3} + 3 x)}^{3}$ $ln(x^3+3x)^3$ ?

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