What is the derivative of $ln (\frac{x^{2} - 4}{2 x + 5})$ $ln((x^2-4)/(2x+5))$ ?

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What is the derivative of $ln (\frac{x^{2} - 4}{2 x + 5})$ $ln((x^2-4)/(2x+5))$ ?

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Answer 1 · 2016-07-30T13:07:51

$\frac{d y}{d x} = \frac{2 x^{2} + 10 x + 8}{(x^{2} - 4) (2 x + 5)}$ $(dy)/(dx) = (2x^2+10x+8)/((x^2-4)(2x+5))$

Explanation:

We have a function y(u(x)) so we need to use the chain rule.

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx) = (dy)/(du)(du)/(dx)$

Set $u = \frac{x^{2} - 4}{2 x + 5}$ $u = (x^2-4)/(2x+5)$ and using the quotient rule obtain:

$\frac{d u}{d x} = \frac{2 x (2 x + 5) - 2 (x^{2} - 4)}{{(2 x + 5)}^{2}} = \frac{2 x^{2} + 10 x + 8}{{(2 x + 5)}^{2}}$ $(du)/(dx) = (2x(2x+5) - 2(x^2-4))/(2x+5)^2 = (2x^2+10x+8)/(2x+5)^2$

$y = ln (u) \Rightarrow \frac{d y}{d u} = \frac{1}{u} = \frac{2 x + 5}{x^{2} - 4}$ $y = ln(u) implies (dy)/(du) = 1/u = (2x+5)/(x^2-4)$

$therefore (dy)/(dx) = (2x^2+10x+8)/((x^2-4)(2x+5))$

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What is the derivative of $ln (\frac{x^{2} - 4}{2 x + 5})$ $ln((x^2-4)/(2x+5))$ ?

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