Given y = (cos^-1x)²
Differentiating with respect to 'x' we get
y' = 2.cos^-1x\(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
\(\sqrt{1-x^2} y_1=-\left(2 \cos ^{-1} x\right)\)
Squaring on both sides
\( \left(1-x^2\right) y_1^2=4\left(\cos ^{-1} x\right)^2 \)
\(\left(1-x^2\right) \dot{y}_1^2=4 y \) (using(1))
Differentiate W. R. T x
\( \left(1-x^2\right)\left(2 y_1 y_2\right)+y_1^2(-2 x)=4 y_1 \)
\(\left(1-x^2\right) 2 y_1 y_2-2 x y_1^2-4 y_1=0\)
Divide it by 2y₁
\(\left(1-x^2\right) y_2-x y_1-2=0\)
when x = 0
\( (1-0) y_2-0 y_1-2=0 \)
\(y_2-2=0 \)
\(y_2=2 \)

\( \sin y =x \sin (a+y) \)
\( x =\frac{\sin y}{\sin (a+y)} \)
\(\frac{d x}{d y} =\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^2(a+y)} \)
\( =\frac{\sin (a+y-y)}{\sin ^2(a+y)} \)
[Since sin(A - B) = sin A cos B - cos A sin B]
\(\frac{d x}{d y}=\frac{\sin a}{\sin ^2(a+y)}\)
Take reciprocal.
\(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}\) \(a \neq n \pi\)
Hence proved.

Given \(y_1=\frac{\sin ^{-1} x}{\sqrt{1-x^2}} .....(1)\)
\(\sqrt{1-x^2} y=\left(\sin ^{-1} x\right)\)
Squaring on both sides, (1 - x²) y² = (sin-1 x)²
Differentiate W.R.T x 
\(\left(1-x^2\right)\left(2 y y_1\right)+y^2(-2 x)=2 \sin ^{-1} x \frac{1}{\sqrt{1-x^2}}\) (using (1))
\(\left(1-x^2\right)\left(2 y y_1\right)-2 x y^2=2 y\)
The above equation divided by 2y.
(1 - x²) y₁ - ay = 1.
Diffrentiate W. R. To x
\( \left(1-x^2\right) y_2+y_1(-2 x)-x y_1-y(1)=0 \)
\(\left(1-x^2\right) y_2-2 x y_1-x y_1-y=0 \)
\(\left(1-x^2\right) y_2-3 x y_1-y=0\)
Hence proved.

Given \(u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)\)
\(u=\tan ^{-1}\left(\frac{2 \sin x / 2 \cos ^x / 2}{2 \cos ^2 x / 2}\right)=\tan ^{-1}(\tan x / 2)\)
\(u =x / 2\)
\( \frac{d u}{d x} =\frac{1}{2} \)
\(v =\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
\(v =\tan ^{-1}\left(\frac{\cos ^2 x / 2-\sin ^2 x / 2}{\left(\sin x / 2+\cos ^x / 2\right)^2}\right)\)
\(v =\tan ^{-1}\left[\frac{(\cos x / 2+\sin x / 2)\left(\cos x / 2-\sin \frac{x}{2}\right)}{(\sin x / 2+\cos x / 2)^2}\right] \)
\(=\tan ^{-1}\left[\frac{\cos x / 2-\sin x / 2}{\sin x / 2+\cos x / 2}\right] \)
\(v =\tan ^{-1}\left[\frac{1-\tan x / 2}{1+\tan x / 2}\right] \)
\(v =\tan ^{-1}[\tan (\pi / 4-x / 2)] \)
\(v =\pi / 4-x / 2 \)
\(\frac{d v}{d x} =-1 / 2 \)
\(\frac{d u}{d v} =\frac{d u / d x}{d v / d x}\)
\(=\frac{1 / 2}{-1 / 2}=-1\)

\( y=\cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right) \)
\( Put\ x=\cos \theta \)
\( \theta=\cos ^{-1} x \)
\( \sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta} / 2}=\sqrt{\tan ^2 \theta / 2}\)
\(=\tan \theta / 2\) \(\left[\because 1-\cos A=2 \sin ^2 A / 2\right.\) \(\left.1+\cos A=2 \cos ^2 A / 2\right]\)
\(y =\cos \left(2 \tan ^{-1}(\tan \theta / 2)\right) \)
\(=\cos (2(\theta / 2))\)
\(=\cos \theta \)
\(=\cos \left(\cos ^{-1} x\right) \)
\(y =x \)
\(\therefore \frac{d y}{d x} =1 \)

\( y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\) \( \left[\because 1-\cos A=2 \sin ^2 A / 2\right.\), \(\left.1+\cos A=2 \cos ^2 A / 2\right]\)

\(=\tan ^{-1} \sqrt{\frac{2 \sin ^2 x / 2}{2 \cos ^2 x / 2}}\)
\(=\tan ^{-1} \cdot \sqrt{\tan ^2 x / 2}=\tan ^{-1}(\tan x / 2)\)
\(y=x / 2\)
\(\frac{d y}{d x}=1 / 2 .\)

We have \(x^2+y^2=4\)
As before, \({dy\over dx}=-{x\over y}\)
Hence, by the quotient rule
\({d^2y\over dx^2}=-{d\over dx}({x\over y})\)
\(=-{y.1-x.{dy\over dx}\over y^2}\)
\(=-{y-x(-{x\over y})\over y^2}\)
\(=-{x^2+y^2\over y^3}=-{4\over y^3}\).

\( \sqrt{x y} =e^{(z-y)} \)
\((x y)^{1 / 2} =e^{x-y}\)
Take log on both sides
\(\frac{1}{2} \log x y=(x-y) \log e\)
\(\frac{1}{2} \log x y=x-y\)
\( \frac{1}{2} \cdot \frac{1}{x y}\left(x \frac{d y}{d x}+y\right) =1-\frac{d y}{d x}\)
\(\frac{1}{2 y} \frac{d y}{d x}+\frac{1}{2 x} =1-\frac{d y}{d x}\)
\(\frac{1}{2 y} \frac{d y}{d x}+\frac{d y}{d x} =1-\frac{1}{2 x}\)
\(\frac{d y}{d x}\left(\frac{1}{2 y}+1\right) =1-\frac{1}{2 x}\)
\(\frac{d y}{d x}=\frac{\frac{2 x-1}{2 x}}{\frac{1+2 y}{2 y}} =\frac{(2 x-1)}{2 x} \cdot \frac{2 y}{(1+2 y)}\)
\(=\frac{y(2 x-1)}{x(1+2 y)}\)

\(y=x^{\cos x}\)
Take log on both sides.
\(\log y =\log \left(x^{\cos x}\right) \)
\( \log y =\cos x \cdot \log x \)
\( \frac{1}{y} \frac{d y}{d x} =\cos x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(\cos x) \)
\( \frac{1}{y} \frac{d y}{d x} =\cos x \cdot \frac{1}{x}+\log x(-\sin x) \)
\( \frac{d y}{d x} =y\left[\frac{\cos x}{x}-\log x(\sin x)\right] \)
\(=x^{\cos x}\left[\frac{\cos x}{x}-\log x(\sin x)\right] \)

tan(x + y) + tan(x - y) = x
Differentiate w.r. to x
\(\sec ^2(x+y)\left[1+\frac{d y}{d x}\right]+\sec ^2(x-y)\left[1-\frac{d y}{d x}\right]=1\)
\( \sec ^2(x+y)+\sec ^2(x+y) \frac{d y}{d x}+\sec ^2(x-y) -\sec ^2(x-y) \frac{d y}{d x}=1\)
\( \frac{d y}{d x}\left[\sec ^2(x+y)-\sec ^2(x-y)\right] =1-\sec ^2(x+y)-\sec ^2(x-y) \)
\(\frac{d y}{d x}=\frac{1-\sec ^2(x+y)-\sec ^2(x-y)}{\sec ^2(x+y)-\sec ^2(x-y)}\)