Hoe do you differentiate $f(x)=ln(e^(4x)+3x)$ ?

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Answer 1 · 2015-11-09T10:29:54

$f'(x)=frac{4e^{4x}+3}{e^{4x}+3x}$

Let $u=e^{4x}+3x$ .

$frac{du}{dx}=4e^{4x}+3$

$f'(x)=frac{d}{dx}[ln(e^{4x}+3x)]$

$=frac{d}{dx}[ln(u)]$

$=frac{d}{du}[ln(u)]frac{du}{dx}$

$=(1/u)(4e^{4x}+3)$

$=frac{4e^{4x}+3}{e^{4x}+3x}$

Hoe do you differentiate f(x)=ln(e^(4x)+3x) f(x)=ln(e^(4x)+3x) ?