Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 50

g(t) = log_2 (t^2+7)^3
Before taking the derivative of the function, apply the logarithm rule log_b (a^m)= m*log_b(a) . So the function becomes:
g(t) = 3log_2(t^2+7)
Take note that the derivative formula of logarithm is d/dx[log_b (u)] = 1/(ln(b)*u)*(du)/dx .
So g'(t) will be:
g'(t) = d/dt [3log_2 (t^2+7)]
g'(t) = 3d/dt [log_2 (t^2+7)]
g'(t) =3 * 1/(ln(2) * (t^2+7)) * d/dt(t^2+7)
g'(t) = 3 * 1/(ln(2) * (t^2+7)) * 2t
g'(t) = (3*2t)/(ln(2) * (t^2+7))
g'(t) = (6t)/((t^2+7)ln(2))

Therefore, the derivative of the function is g'(t) = (6t)/((t^2+7)ln(2)) .