What is the derivative of the function f (x)=ln (ln ((x+4)/ln (x^2+4) ?

1 Answer
Apr 27, 2018

f'(x) = (1/(ln((x+4)/(ln(x^2+4)))))((1)/((x+4))).(((x^2+4)(ln(x^2+4))-(2x^2+4x))/((x^2+4)(ln(x^2+4))))

Explanation:

f'(x) = (1/(ln((x+4)/(ln(x^2+4)))))(1/((x+4)/(ln(x^2+4)))).(((1)(ln(x^2+4))-(x+4)(1)/((x^2+4))(2x))/((ln(x^2+4)))^2)

f'(x) = (1/(ln((x+4)/(ln(x^2+4)))))(ln(x^2+4)/((x+4))).((ln(x^2+4)-(2x^2+4x)/((x^2+4)))/((ln(x^2+4)))^2)

f'(x) = (1/(ln((x+4)/(ln(x^2+4)))))(cancel(ln(x^2+4))/((x+4))).(((x^2+4)(ln(x^2+4))-(2x^2+4x))/((x^2+4)(ln(x^2+4))^cancel(2)))

f'(x) = (1/(ln((x+4)/(ln(x^2+4)))))((1)/((x+4))).(((x^2+4)(ln(x^2+4))-(2x^2+4x))/((x^2+4)(ln(x^2+4))))