`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))` .
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