# Find the derivative of <mrow> ln &#x2061;<!-- ⁡ 3 + 4

Find the derivative of $\frac{\mathrm{ln}\left(3+4x\right)}{6x+5{x}^{2}+2{x}^{3}+2{x}^{4}}$.
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Factor $x$ out of $6x$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot 6+5{x}^{2}+2{x}^{3}+2{x}^{4}}$

Factor $x$ out of $5{x}^{2}$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot 6+x\left(5x\right)+2{x}^{3}+2{x}^{4}}$

Factor $x$ out of $2{x}^{3}$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot 6+x\left(5x\right)+x\left(2{x}^{2}\right)+2{x}^{4}}$

Factor $x$ out of $2{x}^{4}$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot 6+x\left(5x\right)+x\left(2{x}^{2}\right)+x\left(2{x}^{3}\right)}$

Factor $x$ out of $x\cdot 6+x\left(5x\right)$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot \left(6+5x\right)+x\left(2{x}^{2}\right)+x\left(2{x}^{3}\right)}$

Factor $x$ out of $x\cdot \left(6+5x\right)+x\left(2{x}^{2}\right)$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\cdot \left(6+5x+2{x}^{2}\right)+x\left(2{x}^{3}\right)}$

Factor $x$ out of $x\cdot \left(6+5x+2{x}^{2}\right)+x\left(2{x}^{3}\right)$.

$\frac{\mathrm{ln}\left(3+4x\right)}{x\left(6+5x+2{x}^{2}+2{x}^{3}\right)}$

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