Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\mathrm{e}^t(t^3+t^2)\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=t^3\mathrm{e}^t(t^2+4t+2)\).

Since \(x\) and \(y\) are both given in terms of \(t\), use parametric differentiation:

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\).

(a) Differentiate with respect to \(t\):

From \(y=t\mathrm{e}^t\),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{e}^t+t\mathrm{e}^t=\mathrm{e}^t(t+1)\).

From \(x=1+\frac{1}{t}=1+t^{-1}\),

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2}\).

Hence

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{e}^t(t+1)}{-t^{-2}}=-t^2\mathrm{e}^t(t+1)=-\mathrm{e}^t(t^3+t^2)\).

(b) For the second derivative, use

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}\).

From part (a),

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\mathrm{e}^t(t^3+t^2)\).

Differentiate with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\mathrm{e}^t(3t^2+2t)-\mathrm{e}^t(t^3+t^2)=-\mathrm{e}^t(t^3+4t^2+2t)\).

Also, since \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2}\), we have

\(\displaystyle \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2\).

Therefore

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\mathrm{e}^t(t^3+4t^2+2t)(-t^2)=t^2\mathrm{e}^t(t^3+4t^2+2t)\).

Factorising,

\(\displaystyle t^2\mathrm{e}^t(t^3+4t^2+2t)=t^3\mathrm{e}^t(t^2+4t+2)\).

So \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=t^3\mathrm{e}^t(t^2+4t+2)\).