Question 1088613
The slope(gradient) of a curve is equal to the value of the derivative at that point.
So using the chain rule,
{{{dx/dt=(2t-5)3(t+1)^2+(t+1)^3(2)}}}
{{{dx/dt=(t+1)^2(3(2t-5)+2(t+1))}}}
{{{dx/dt=(t+1)^2(6t-15+2t+2)}}}
{{{dx/dt=(t+1)^2(8t-13)}}}
So then when {{{t=3}}},
{{{m=dx/dt=(3+1)^2(8(3)-13)}}}
Work that out for your answer.