\( y =\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
\( \text { Take } x =\tan \theta \Rightarrow \theta=\tan ^{-1} x \)
\( \therefore \frac{1-x^2}{1+x^2} =\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta \)
\( \therefore y =\cos ^{-1}(\cos 2 \theta) \)
\( y =2 \theta \)
\( y =2 \tan ^{-1} x \)
\( \frac{d y}{d x} =\frac{2}{1+x^2} \)

\( \frac{x^2}{a^2}+\frac{y^2}{b^2} =1\)
\(\frac{2 x}{a^2}+\frac{2 y}{b^2} \cdot \frac{d y}{d x} =0 \)
\(\frac{2 y}{b^2} \frac{d y}{d x} =\frac{-2 x}{a^2}\)
\(\frac{d y}{d x}=\frac{-2 x}{a^2} \times \frac{b^2}{2 y} =\frac{-b^2 x}{a^2 y}\)

Take u = x^x,v = x log x
log u = x log x
\({1\over u}{du \over dx}=x.{1\over x}+1.log \ x=1+log \ x\)
\({du \over dx}=u({1}+log \ x)=x^x(1+log \ x)\)
\(\frac{d v}{d x}=1+\log x\)
\(\frac{d\left(x^x\right)}{d(x \log x)}=\frac{d u}{d V}=\frac{\frac{d u}{d x}}{\frac{d V}{d x}}=x^X\)

Let x = cos \(\theta\)
Then 4x³ − 3x = 4cos³ \(\theta\)- 3cos = cos3\(\theta\) and
f(x) = cos^-1(cos3\(\theta\) ) = 3\(\theta\) = 3cos^-1x
Therefore,f'(x) = 3\(({-1\over \sqrt{1-x^2}})={-3\over \sqrt{1-x^2}}.\)

Given y = \(y=500{ e }^{ 7x }+600{ e }^{ -7x }\)  ....(1)
Differentiating both sides with respect to 'x' we have
\(\frac { dy }{ dx } =500{ e }^{ 7x }(7)+600{ e }^{ -7x }(-7)\)
\( \frac { dy }{ dx } =7(500{ e }^{ 7x }-600{ e }^{ -7x })\)
Differentiating again with respect to 'x' we have
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =7(500{ (e }^{ 7x })(7)-600{ (e }^{ -7x })(-7))\)
\(=49(500{ \quad e }^{ 7x }+600{ \quad e }^{ -7x })=\quad 49\quad y\quad (from\quad (1))\)
\(\therefore \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =49y\)  Hence proved.

\(x^y=y^x\)
Take log on both sides
y log x = x log y
\( \frac{y}{x} \div \log x \frac{d y}{d x}=\frac{x d y}{y d x}+\log y \)
\(\log x \frac{d y}{d x}-\frac{x d y}{y d x}=\log y-\frac{y}{x} \)
\(\frac{d y}{d x}\left(\log x-\frac{x}{y}\right)=\frac{x \log y-y}{x} \)
\(\frac{d y}{d x}=\frac{x \log y-y}{x}=\frac{y(x \log y-y)}{x(y \log x-x)}\)

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

We have, f(x) = x cos x.
Now f'(x) = - x sin x + cos x, and
f'(x) = - (x cos x + sin x) - sin x
= - x cos x - 2 sin x.

We have, y = x³ - 6x² - 5x + 3 and
y' = 3x² - 12x - 5 
y'' = 6x - 12
y''' = 6.

We have x = at² ; y = 2at
\({dy\over dx}={y'(t)\over x'(t)}={2a\over 2at}={1\over t}.\)

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 x = a(θ + sin θ), y = a(1 - cosθ)
\( \frac{d x}{d \theta}=a(1+\cos \theta) \quad \frac{d y}{d \theta}=a(\sin \theta) \)
\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{a \sin \theta}{a(1+\cos \theta)} \)
\(=\frac{2 \sin \theta / 2 \cos \theta / 2}{2 \cos ^2 \theta / 2}=\frac{\sin \theta / 2}{\cos \theta / 2}\)
\( y^{\prime} =\tan \theta / 2 \)
\( y^{\prime \prime} =\sec ^2 \theta / 2\left(\frac{1}{2}\right) \)
At \( \theta=\pi / 2 \quad y^{\prime \prime} =\sec ^2 \pi / 4\left(\frac{1}{2}\right) \)
\( =(\sqrt{2})^2 \cdot \frac{1}{2} \)
\( =2\left(\frac{1}{2}\right)=1 \)

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.

y = e^{tan-1 x} ...........(1)
⇒ y'=\({ e }^{ tan^{ -1 }x }(\frac { 1 }{ 1+{ x }^{ 2 } }) \)
\(\left(1+x^2\right) y^{\prime}=e^{\tan ^{-1} x} \)
\(\left(1+x^2\right) y^{\prime}=y\) (using (1))
Again Diff w. r. to x.
(1 + x²) y" + y' (2x) = y'
(1 + x²) y" + y' (2x) - y' = 0
(1 + x²) y" + (2x - 1) y' = 0
Hence proved.

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

\( v=\tan ^{-1} x \Rightarrow \frac{d v}{d x}=\frac{1}{1+x^2} \)
\(u=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x} \)
\(Put\ x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\)
\(W.K.T 1+\tan ^2 \theta=\sec ^2 \theta\)
\(\sec ^2 \theta-1=\tan ^2 \theta\).
\(\therefore u =\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \)
\(=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\sin \theta / \cos \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta / \cos \theta}{\sin \theta / \cos \theta}\right) \)
\(=\tan ^{-1}\left(\frac{2 \sin ^2 \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}\right)=\tan ^{-1}(\tan \theta / 2) \)
\( u =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x \text { (using (1)) }\)
\( \frac{d u}{d x} =\frac{1}{2}\left(\frac{1}{1+x^2}\right) \)
\( \therefore \frac{d u}{d v} =\frac{d u / d x}{d v / d x}=\frac{1 / 2\left(\frac{1}{1+x^2}\right)}{\left(1 / 1+x^2\right)} \)
\(\frac{d u}{d v} =1 / 2 \)