Can someone help me solving the following problem

Let V(t) = 2ln(t).

(i) Find the slope of the function at t = 1.
Interpret your answer.
(ii) Find the equation of the tangent at t = 1.

For this, recall that the derivative of log(t) (ln(t) ) is 1/t

Hence the slope of V at t is (2/t) so at 1 the slope is 2.

V(1)=2log(1)=0 (Log(1)=0)

Hence the tangent line has (cartesian) equation y=2x-2 ( -2 since y(1)=V(1)=0)

2ln=G(V)