How do you differentiate $f(x)=ln(e^(x^2+1)/(2x^3-1)^(1/2))$ ?

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Answer 1 · 2015-11-28T21:41:19

$f'(x)= (x(4x^3 -3x-2))/(2x^3 -1)$

This problem look intimidating but if we can rewrite it using logarithm properties, it actually really easy.

*Remember: $color(red)(log(a/b) hArr loga- log b$ ; $color(blue)(loga^n = n log a$ ; $color(green)(ln e= 1)$

Step 1: Rewrite the logarithm expression
$f(x)= ln((e^(x^2+1))/(2x^3-1)^(1/2))$ $hArr color(red)(f(x)= ln(e^(x^2+1))-ln(2x^3-1)^(1/2)$

$hArr f(x)= color(blue)((x^2 +1)*(ln e))-1/2ln (2x^3-1)$

$hArr f(x)=color(green)( (x^2 +1)) -1/2 ln(2x^3 -1)$

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Step 2: Now can can begin to differentiate the function
*Remember $color(red)(g(x)= lnx ; g'(x)= (dx)/x)$

$f'(x)= color(green)(2x) -1/2((color(red)(6x^2))/(2x^3-1))$ Differentiate

$f'(x)= color(green)(2x) -((color(red)(3x^2))/(2x^3-1))$ Simplify

$f'(x)= (color(green)(2x)color(red)( (2x^3 -1)) -3x^2)/(2x^3 -1)$ Common denominator

$f'(x)= (4x^4 -2x-3x^2)/(2x^3 -1)$ Distributen and simplify

$f'(x)= (x(4x^3 -3x-2))/(2x^3 -1)$ Better answer :)

How do you differentiate f(x)=ln(e^(x^2+1)/(2x^3-1)^(1/2)) f(x)=ln(e^(x^2+1)/(2x^3-1)^(1/2)) ?