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

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

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Answer 1 · 2016-10-19T00:57:21

$\frac{d}{d x} ln (x^{3}) = \frac{3}{x}$ $d/dxln(x^3)=3/x$

$\frac{d^{2}}{{d x}^{2}} ln (x^{3}) = - \frac{3}{x^{2}}$ $(d^2)/(dx^2)ln(x^3)=-3/x^2$

Explanation:

Using the chain rule, the power rule, and the product rule, along with the derivative $\frac{d}{d x} ln (x) = \frac{1}{x}$ $d/dx ln(x) = 1/x$ , we have

First Derivative:

$\frac{d}{d x} ln (x^{3}) = \frac{1}{x^{3}} (\frac{d}{d x} x^{3})$ $d/dxln(x^3) = 1/x^3(d/dxx^3)$

$= \frac{1}{x^{3}} (3 x^{2})$ $=1/x^3(3x^2)$

$= \frac{3}{x}$ $=3/x$

Second Derivative:

$\frac{d^{2}}{{d x}^{2}} ln (x^{3}) = \frac{d}{d x} (\frac{d}{d x} ln (x^{3}))$ $(d^2)/(dx^2)ln(x^3) = d/dx(d/dxln(x^3))$

$= \frac{d}{d x} (\frac{3}{x})$ $=d/dx(3/x)$

$= \frac{d}{d x} 3 x^{- 1}$ $=d/dx3x^(-1)$

$= 3 (- x^{- 2})$ $=3(-x^-2)$

$= - \frac{3}{x^{2}}$ $=-3/x^2$

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

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