The tangent line has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], which we can find using the chain rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
We have
[tex]x(t)=3t^2+7\implies\dfrac{\mathrm dx}{\mathrm dt}=6t[/tex]
[tex]y(t)=t^6\implies\dfrac{\mathrm dy}{\mathrm dt}=6t^5[/tex]
so that
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{6t^5}{6t}=t^4[/tex]
When [tex]t=-1[/tex], the tangent line has slope 1.
To compute the second derivative [tex]\frac{\mathrm d^2y}{\mathrm dx^2}[/tex], first notice that the first derivative [tex]\frac{\mathrm dy}{\mathrm dx}[/tex] turns out to be a function of [tex]t[/tex]. Denote this first derivative by [tex]f(t)[/tex]. Then by the chain rule, we find
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{4t^3}{6t}=\dfrac23t^2[/tex]
and at [tex]t=1[/tex], we get a value of 2/3.
