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

Question

Luke Alvarez · Accepted Answer

#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 :)

Cameron Alexander · Answer

undefined To differentiate $ f(x) = \ln\left(\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}\right) $, we use the chain rule and the quotient rule. The derivative is:

$$ f'(x) = \frac{1}{\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}} \cdot \left(\frac{d}{dx}\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}\right) $$

$$ f'(x) = \frac{1}{\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}} \cdot \left(\frac{(2x)(e^{x^2+1})(2x^3-1)^{-1/2} - e^{x^2+1}(1/2)(2x^3-1)^{-3/2}(6x^2)}{(2x^3-1)}\right) $$

$$ f'(x) = \frac{(2x)(2x^3-1)^{1/2} - 3x(2x^3-1)}{x^2+1} \cdot \frac{1}{(2x^3-1)^{1/2}} $$