Question

Find the second derivative of the function.g(t) = 31t-2g"(t) =X

Answer 1

Let's find the first derivative of the function:

`g^(\prime)(t)=(dg)/(dt)=(d)/(dt)(31t^(-2))=-2(31)t^(-3)=-62t^(-3)`

Hence the first derivative is:

`(dg)/(dt)=-62t^(-3)`

Now, that we have the first derivative we can calculate the second derivative:

`g^(\doubleprime)(t)=(d^2g)/(dt^2)=(d)/(dt)(-62t^(-3))=-3(-62)t^(-4)=186t^(-4)`

Therefore:

`g^(\doubleprime)(t)=186t^(-4)`