How do you find the first and second derivative of # (ln(x^(2)+3))^(3)#?

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Answer 1 · 2017-01-31T14:15:26

#f'(x)=(6xln^2(x^2+3))/(x^2+3)#

#f''(x)=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#

The first derivative is:

#f'(x)=3ln^2(x^2+3)*1/(x^2+3)*2x#

#=(6xln^2(x^2+3))/(x^2+3)#

The second derivative is:

#f''(x)=((6ln^2(x^2+3) + 6x*2 ln(x^2+3)* 1/(x^2+3) * 2x)*(x^2+3)-6xln^2(x^2+3)*2x)/(x^2+3)^2#

#=((6ln^2(x^2+3) + (24x ln(x^2+3))/(x^2+3))*(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=(((6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3))/cancel(x^2+3))*cancel((x^2+3))-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=(6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=((6x^2+18-12x^2)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#

#=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#