Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{t^2+1+\sqrt{t^2+1}}{t^2}\).

(b) When \(t=\frac34\), \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{80}{3}\).

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

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-1+\frac{1}{1+t^2}, \qquad \frac{\mathrm{d}y}{\mathrm{d}t}=1+\frac{1}{\sqrt{1+t^2}}.\)

Now

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

Simplify each derivative:

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-1+\frac{1}{1+t^2}=-\frac{t^2}{1+t^2},\)

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=1+\frac{1}{\sqrt{1+t^2}}=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}}.\)

Hence

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

Since \(1+t^2=(\sqrt{1+t^2})^2\),

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

(b) Let

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

Write \(N=1+t^2+\sqrt{1+t^2}\), so that \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{N}{t^2}\).

Then

\(\displaystyle N'=2t+\frac{t}{\sqrt{1+t^2}}.\)

Using the quotient rule,

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\frac{t^2N'-2tN}{t^4}.\)

So

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-\frac{t^2\left(2t+\frac{t}{\sqrt{1+t^2}}\right)-2t\left(1+t^2+\sqrt{1+t^2}\right)}{t^4}.\)

This simplifies to

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

For parametric equations,

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

Using \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{t^2}{1+t^2}\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{t^2(1+t^2)^{-1/2}-2(1+\sqrt{1+t^2})}{t^5}(1+t^2).\)

Now set \(t=\frac34\). Then

\(\displaystyle 1+t^2=\frac{25}{16}, \qquad \sqrt{1+t^2}=\frac54, \qquad t^5=\left(\frac34\right)^5=\frac{243}{1024}.\)

Also,

\(\displaystyle t^2(1+t^2)^{-1/2}=\frac{9}{16}\cdot\frac45=\frac{9}{20},\)

\(\displaystyle 2(1+\sqrt{1+t^2})=2\left(1+\frac54\right)=\frac92.\)

So

\(\displaystyle \frac{t^2(1+t^2)^{-1/2}-2(1+\sqrt{1+t^2})}{t^5}(1+t^2)=\left(\frac{9}{20}-\frac92\right)\cdot\frac{25}{16}\cdot\frac{1024}{243}.\)

Since \(\displaystyle \frac{9}{20}-\frac92=-\frac{81}{20}\), this gives

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{81}{20}\cdot\frac{25}{16}\cdot\frac{1024}{243}=-\frac{80}{3}.\)