What is the derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ ?

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What is the derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ ?

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Answer 1 · 2018-04-05T11:10:28

$\frac{d}{d x} ln (e^{4 x} + 3 x) = \frac{4 e^{4 x} + 3}{e^{4 x} + 3 x}$ $d/(dx) ln(e^(4x)+3x)=(4e^(4x)+3)/(e^(4x)+3x)$

Explanation:

Derivative of $ln x$ $lnx$ is $\frac{1}{x}$ $1/x$

So derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ is $\frac{1}{e^{4 x} + 3 x} \frac{d}{d x} (e^{4 x} + 3 x)$ $1/(e^(4x)+3x)d/dx(e^(4x)+3x)$ (Chain rule)

Derivative of $e^{4 x} + 3 x$ $e^(4x)+3x$ is $4 e^{4 x} + 3$ $4e^(4x)+3$

So derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ is $\frac{1}{e^{4 x} + 3 x} \cdot (4 e^{4 x} + 3)$ $1/(e^(4x)+3x)*(4e^(4x)+3)$

$= \frac{4 e^{4 x} + 3}{e^{4 x} + 3 x}$ $=(4e^(4x)+3)/(e^(4x)+3x)$

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What is the derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ ?

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What is the derivative of $ln (e^{4 x} + 3 x)$ $ln(e^(4x)+3x)$ ?