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

1 Answer
Jul 12, 2017

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = (3x^5-x^3-3x^2+1)/(x(x^2+1)(2x^3-1))#

Explanation:

Note that in general, using the chain rule:

(1) #d/dx ln(f(x)) = (f'(x))/f(x)#

Using the properties of logarithms we have:

#ln ( (x (x^2+1))/(2x^3-1)^(1/2)) = lnx + ln(x^2+1) -1/2ln(2x^3-1)#

and as the derivative is linear:

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = d/dx lnx + d/dx ln(x^2+1) -1/2d/dx ln(2x^3-1)#

Then, based on (1):

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = 1/x + (2x)/(x^2+1) -1/2(6x^2)/(2x^3-1)#

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = 1/x + (2x)/(x^2+1) -(3x^2)/(2x^3-1)#

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = ((2x^3-1)(x^2+1)+2x^2(2x^3-1)-3x^3(x^2+1))/(x(x^2+1)(2x^3-1))#

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = (2x^5-x^2+2x^3+1+4x^5-2x^2-3x^5-3x^3)/(x(x^2+1)(2x^3-1))#

#d/dx (ln ( (x (x^2+1))/(2x^3-1)^(1/2)) ) = (3x^5-x^3-3x^2+1)/(x(x^2+1)(2x^3-1))#