By the chain rule,
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}[/tex]
Then for all [tex]t[/tex] the first derivative has a value of 7.
By the product rule,
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\right]=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}\dfrac{\mathrm dt}{\mathrm dx}+\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm d^2t}{\mathrm dx^2}[/tex]
but [tex]t=x\implies\dfrac{\mathrm dt}{\mathrm dx}=1\implies\dfrac{\mathrm d^2t}{\mathrm dx^2}=0[/tex], so we're left with
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}[/tex]
By the chain rule,
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\mathrm d^2y}{\mathrm dt^2}[/tex]
but [tex]y=7t-2\implies\dfrac{\mathrm dy}{\mathrm dt}=7\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=0[/tex] so the second derivative is 0 for all [tex]t[/tex].
