What is the derivative of $ln (2 x + 1)$ $ln(2x+1)$ ?

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What is the derivative of $ln (2 x + 1)$ $ln(2x+1)$ ?

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Answer 1 · 2016-10-18T20:05:08

$\frac{2}{2 x + 1}$ $2/(2x+1)$

Explanation:

$y = ln (2 x + 1)$ $y=ln(2x+1)$ contains a function within a function, i.e. $2 x + 1$ $2x+1$ within $ln (u)$ $ln(u)$ . Letting $u = 2 x + 1$ $u=2x+1$ , we can apply chain rule.

Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

$\frac{d y}{d u} = \frac{d}{d u} ln (u) = \frac{1}{u}$ $(dy)/(du)=d/(du)ln(u)=1/u$

$\frac{d u}{d x} = \frac{d}{d x} 2 x + 1 = 2$ $(du)/(dx)=d/(dx)2x+1=2$

$:.(dy)/(dx)=1/u*2=1/(2x+1)*2=2/(2x+1)$

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What is the derivative of $ln (2 x + 1)$ $ln(2x+1)$ ?

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