Find the derivative of the function. h(t) = (t + 5)^ 2/3 (2t^2 − 4)^3

Answer 1

product rule says:

(t+5)^(2/3) *3 *( 2t^2 -4)^2 ( 4t) + (2/3)(t+5)^(-1/3) (2t^2 -4)^3

(t+5)^(-1/3) (2t^2-4)^2 [ 3*(t+5)*(4t) + (2/3)(2t^2-4) ]=

(t+5)^(-1/3) (2t^2-4)^2 [ 12t^2 + 60t + (4/3) t^2 - 8/3 ] =

(t+5)^(-1/3) (2t^2-4)^2 [ (36/3)t^2 + 60t + (4/3) t^2 - 8/3] =

(t+5)^(-1/3) (2t^2-4)^2 [ (40/3)t^2 + 60t - 8/3] =

(t+5)^(-1/2) [2(t^2-2)]^2 (4/3) [ 10t^2 - 45t - 2] =

(t+5)^(-1/2) 4(4/3) (t^2-2)(10t^2 - 45t - 2] =

(16/3)(t^2-2)(10t^2 - 45t - 2] / (t+5)^(1/2)