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

1 Answer
Nov 28, 2015

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

Explanation:

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 :)