What is the derivative of $\frac{2 ln (x + 3)}{x^{2}}$ $(2ln(x+3))/x^2$ ?

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What is the derivative of $\frac{2 ln (x + 3)}{x^{2}}$ $(2ln(x+3))/x^2$ ?

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Answer 1 · 2018-04-01T16:29:40

answer = $\frac{-2(x+6)}{x^{3}}$ $"-2(x+6)"/x^3$

Explanation:

$\frac{d}{d x} (\frac{u}{v}) = \frac{u'v-v'u}{v^{2}}$ $d/dx(u/v)="u'v-v'u"/v^2$
similarly, $\frac{d}{d x} (\frac{2 (x + 3)}{x^{2}}) = \frac{d}{d x} (\frac{2 x + 6}{x^{2}})$ $d/dx((2(x+3))/x^2)=d/dx((2x+6)/x^2)$
= $\frac{(x^{2}) \frac{d}{d x} (2 x + 6) - (2 x + 6) \frac{d}{d x} (x^{2})}{{(x^{2})}^{2}}$ $((x^2)d/dx(2x+6)-(2x+6)d/dx(x^2))/(x^2)^2$
= $\frac{(x^{2}) (2) - (2 x + 6) (2 x)}{x^{4}}$ $((x^2)(2)-(2x+6)(2x))/x^4$
= $\frac{2 x^{2} - 4 x^{2} - 12 x}{x^{4}}$ $(2x^2-4x^2-12x)/x^4$
$\frac{- 2 x^{2} - 12 x}{x^{4}}$ $(-2x^2-12x)/x^4$
$\frac{- 2 x (x + 6)}{x^{4}} = \frac{- 2 (x + 6)}{x^{3}}$ $(-2x(x+6))/x^4=(-2(x+6))/x^3$

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