How do you differentiate $ln (x + 4 + e^{- 3} x)$ $ln(x+4+e^-3x)$ ?

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Answer 1 · 2018-03-14T13:24:45

$\frac{1 - 3 e^{- 3 x}}{x + 4 + e^{- 3 x}}$ $color(blue)((1-3e^(-3x))/(x+4+e^(-3x)))$

If:

$y = ln (x) \Leftrightarrow e^{y} = x$ $y=ln(x)<=>e^y=x$

Using this definition for the given function:

$e^{y} = x + 4 + e^{- 3 x}$ $e^y=x+4+e^(-3x)$

Differentiating implicitly:

$e^{y} \frac{d y}{d x} = 1 + 0 - 3 e^{- 3 x}$ $e^ydy/dx=1+0-3e^(-3x)$

Dividing by: $88 e^{y}$ $color(white)(88)bb(e^y)$

$\frac{d y}{d x} = \frac{1 - 3 e^{- 3 x}}{e^{y}}$ $dy/dx=(1-3e^(-3x))/e^y$

From above:

$e^{y} = x + 4 + e^{- 3 x}$ $e^y=x+4+e^(-3x)$

$:.$

$dy/dx=color(blue)((1-3e^(-3x))/(x+4+e^(-3x)))$

How do you differentiate ln(x+4+e^-3x)ln(x+4+e−3x) ln(x+4+e^-3x)?