Question #01dcb

1 Answer
Oct 1, 2017

= (-4cos(x)sin(x))/(1+3cos^2x)

Explanation:

d/dx(2/3 ln(1 + 3cos^2x)) = 2/3 * (d/dx (ln(1+3cos^2x)))

(constant pulls out of derivative)

knowing that d/dx(ln(u)) = 1/u * u',

= 2/3 * 1/ (1+3cos^2x)*d/dx(1+3cos^2x)

= 2/(3(1+3cos^2x))*(0 + 3⋅(2cos(x))⋅(-sin(x)))

= 2/(3+9cos^2x)*(-6cos(x)sin(x)) = (-12cos(x)sin(x))/(3+9cos^2x)

= (-4cos(x)sin(x))/(1+3cos^2x)