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

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Answer 1 · 2017-07-12T23:19:23

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

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):

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

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

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

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

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

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