Saturday, August 30, 2014

`g(t) = log_2(t^2+7)^3` Find the derivative of the function

`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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