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

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Answer 1 · 2016-10-17T12:00:38

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

We us the chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $dy/dx=dy/(du)(du)/dx$

Let $y = ln (x^{3} + 3 x)$ $y=ln(x^3+3x)$ , and
Let $u = x^{3} + 3 x$ $u=x^3+3x$

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

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

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

How do you find the derivative of ln(x^3+3x)ln(x3+3x)ln(x^3+3x)?