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

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

Given \(x=\sin^{-1}t\) and \(y=t\cos^{-1}t\), where \(0\le t\lt 1\).

(a)

Differentiate with respect to \(t\):

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

For \(y=t\cos^{-1}t\), use the product rule:

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

Hence

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

So

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

(b)

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

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

Using the product rule,

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

Therefore

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

Now

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

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

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

Hence

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