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

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-t^3(1-t^2)^{-\frac32}(2-t^2)\).

Hence \(a=3\) and \(b=-\frac32\).

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

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

Now

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

Also

\(\displaystyle x=1+\frac1t \implies \frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{1}{t^2},\)

so

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

Therefore

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

(b) Differentiate \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t^2}{\sqrt{1-t^2}}\) with respect to \(t\):

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

This can be written as

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

Now use

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

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

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

So

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-2t^3(1-t^2)^{-1/2}-t^5(1-t^2)^{-3/2}.\)

Factorising gives

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

Hence

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

Comparing with \(\,-t^a(1-t^2)^b(2-t^2)\), we get

\(\displaystyle a=3,\quad b=-\frac32.\)