\(x^y+y^x=1\)
Let \(u=x^y, v=y^x\)
Hence,
u+v=1
Differentiating both sides w.r.t. x.
\(\frac{d(v+u)}{d x}=\frac{d(1)}{d x} \)
\(\frac{d v}{d x}+\frac{d u}{d x}=0\)
(Derivative of constant is 0 )

\(We\quad have:\quad y=3{ e }^{ 2x }+{ 2e }^{ 3x }\)
\(\frac { dy }{ dx } =6{ e }^{ 2x }+6{ e }^{ 3x }=6({ e }^{ 2x }+{ e }^{ 3x })\)
\(and\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =12{ e }^{ 2x }+18{ e }^{ 3x }=6(2{ e }^{ 2x }+3{ e }^{ 3x })\)
\(Now\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ { dx }^{ 2 } } +6y\)
\(=6(2{ e }^{ 2x }+3{ e }^{ 3x })-5.6({ e }^{ 2x }+{ e }^{ 3x })+6(3{ e }^{ 2x }+{ 2e }^{ 3x })\)

\(Let\quad y=2\sqrt { cot\quad { x }^{ 2 } } \)
Put x² = t, cot t = s.,
\(y=2\sqrt { s } \)
\(\frac { dy }{ ds } =2.\frac { 1 }{ 2\sqrt { s } } =\frac { 1 }{ \sqrt { s } } \)
\(\frac { ds }{ dt } =-{ { { cosec }^{ 2 } }t },\quad \frac { dt }{ dx } =2x\)
\(\frac { dy }{ dx } =\frac { dy }{ ds } .\frac { ds }{ dt } .\frac { dt }{ dx } \)
\(=\frac { 1 }{ \sqrt { s } } (-{ { cosec }^{ 2 } }t).(2x)\)
\(=\frac { -2x }{ \sqrt { cot\quad x } } { { cosec }^{ 2 } }t\)
\(=\frac { -2x }{ \sqrt { { cot\quad }x^{ 2 } } } { cosec }^{ 2 }\quad { x }^{ 2 }\)

\(Here\quad y={ sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } \)
\(={ sin }^{ -1 }\left( \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
\(={ sin }^{ -1 }(sin2\theta )=2\theta =2{ tan }^{ -1 }x\)
\(Hence,\quad \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \)

\(Let\quad y=cos(log\quad x+{ e }^{ x })\)
\(\frac { dy }{ dx } =-sin(log\quad x+{ e }^{ x }).\frac { d }{ dx } (log\quad x+{ e }^{ x })\)
\(=-sin(log\quad x+{ e }^{ x })\left( \frac { 1 }{ x } +{ e }^{ x } \right) \)
\(=-\frac { 1 }{ x } sin(log\quad x+{ e }^{ x })(1+{ xe }^{ x }),x>0\)

\(Let\quad y={ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(Taking\quad logs.,\quad log\quad y=log{ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(\\ =2log(x+3)+3log(x+4)+4log(x+5)\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ y } \frac { dy }{ dx } \)
\(=\frac { 2 }{ x+3 } (1+0)+\frac { 3 }{ x+4 } (1+0)+\frac { 4 }{ x+5 } (1+0)\)
\(\Rightarrow \frac { dy }{ dx } =y\left[ \frac { 2 }{ x+3 } +\frac { 3 }{ x+4 } +\frac { 4 }{ x+5 } \right] \)
\(Hence,\quad \frac { dy }{ dx } ={ \left( x+3 \right) }^{ 2 }{ \left( x+4 \right) }^{ 3 }{ \left( x+5 \right) }^{ 4 }\)
\(=\left[ \frac { 2 }{ x+3 } +\frac { 3 }{ x+4 } +\frac { 4 }{ x+5 } \right] \)

\(We\quad have:\)
\(X=\sqrt { { a }^{ { sin }^{ -1 }t } } and\quad y=\sqrt { { a }^{ { cos }^{ -1 }t } } \)
\(i.e.,x={ \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }and\quad y={ \left( { \alpha }^{ { cos }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }\)
\(\frac { dx }{ dt } =\frac { 1 }{ 2 } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ -\frac { 1 }{ 2 } }\frac { d }{ dt } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }\)
\(=\frac { 1 }{ 2 } \frac { 1 }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } { \alpha }^{ { sin }^{ -1 }t }{ log }_{ e }\alpha \frac { d }{ dx } \left( { sin }^{ -1 }t \right) \)
\(=\frac { { log }_{ e }\alpha }{ 2 } .\sqrt { { \alpha }^{ { sin }^{ -1 }t } } .\frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(Similarly\quad \frac { dy }{ dt } =-\frac { { log }_{ e }\alpha }{ 2 } \sqrt { { \alpha }^{ { cos }^{ -1 }t } } \frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { \sqrt { { \alpha }^{ { cos }^{ -1 }t } } }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } =-\frac { y }{ x } .,\quad which\quad is\quad true.\)

\(We\quad have:\quad y=3\quad cos(log\quad x)+4\quad sin(log\quad x)....(1)\)
\(Diff.\quad w.r.t.\quad x,{ y }_{ 1 }=-3\quad sin(log\quad x).\frac { 1 }{ x } +4cos(log\quad x).\frac { 1 }{ x } \)
\({ xy }_{ 1 }=-3\quad sin(log\quad x)+4\quad cos(log\quad x)....(2)\)
\(Again\quad diff.\quad w.r.t.\quad x,\)
\({ xy }_{ 2 }+{ y }_{ 1 }=-3\quad cos(log\quad x).\frac { 1 }{ x } -4\quad sin\quad (log\quad x).\frac { 1 }{ x } \)
\({ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }=-(3\quad cos(log\quad x)+4sin(log\quad x))\)
\(Hence,{ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }+y=0\)

\(We\quad have:\quad y=500{ e }^{ 7x }+600{ e }^{ -7x }\)
\(\frac { dy }{ dx } =500{ e }^{ 7x }(7)+600{ e }^{ -7x }(-7)\)
\(=3500{ e }^{ 7x }-4200{ e }^{ -7x }\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =3500{ e }^{ 7x }(7)-4200{ e }^{ -7x }(-7)\)
\(=49(500{ e }^{ 7x }+600{ e }^{ -7x })\)
\(Hence,\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =49y\)

\(We\quad have:\quad y={ \left( { tan }^{ -1 }x \right) }^{ 2 }\)
\(Diff.w.r.t.x,\frac { dy }{ dx } =2\left( { tan }^{ -1 }x \right) .\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } =2\left( { tan }^{ -1 }x \right) \)
\(Squaring{ \left( 1+{ x }^{ 2 } \right) }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }=4{ \left( { tan }^{ -1 }x \right) }^{ 2 }\)
\({ \left( { x }^{ 2 }+1 \right) }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }=4y\)
\(Again\quad diff.w.r.t.\quad x,\quad { \left( { x }^{ 2 }+1 \right) }^{ 2 }2\left( \frac { dy }{ dx } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)
\(+2\left( { x }^{ 2 }+1 \right) 2x{ \left( \frac { dy }{ dx } \right) }^{ 2 }=4\frac { dy }{ dx } .\)
\(Dividing\quad by\quad 2\frac { dy }{ dx } ,\quad we\quad get:\)
\({ \left( { x }^{ 2 }+1 \right) }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +2x\left( { x }^{ 2 }+1 \right) \frac { dy }{ dx } =2\)
\(\Rightarrow { \left( { x }^{ 2 }+1 \right) }^{ 2 }{ y }_{ 2 }+2x\left( { x }^{ 2 }+1 \right) { y }_{ 1 }=2,\quad which\quad is\quad true.\)

\(y=(\sin x)^{x}+\sin^{-1}\sqrt{x}\)
y=e^{x \log(\sin x)}+\sin^{-1}\sqrt{x}
Diff. w.r.t x
\(\frac{dy}{dx}=e^{x \log(\sin x)}[\log\sin x+x\frac{\cos x}{\sin x}]+\frac{1}{\sqrt{1-x}}\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{dy}{dx}=(\sin x)^{x}[\log\sin x+x\cot x]+\frac{1}{2\sqrt{x-x^{-2}}}\)

xⁿyⁿ = (x+y)^m+n 
Taking log we get
logx^m + logyⁿ = (m + n) log (x + y)
mlog x + logy = (m + n)log(x + y)
Differentiating w.r. to x, we get 
\(\frac { m }{ x } +\frac { n }{ y } \frac { dy }{ dx } =\frac { m+n }{ x+y } \times \left( 1+\frac { dy }{ dx } \right) \)
\(\Rightarrow \left( \frac { n }{ y } -\frac { m+n }{ x+y } \right) \frac { dy }{ dx } =\frac { x(m+n)-m(x+y) }{ x\left( x+y \right) } \)
\(\left( \frac { nx+ny-my-ny }{ y(x+y) } \right) \frac { dy }{ dx } =\frac { xm+xn-mx-my }{ x\left( x+y \right) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } \times \left( \frac { xn-my }{ xn-my } \right) \)
= \(\frac { y }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } \)

We have \(\sin (x+y)=x^{2}+y^{2}\) 
On differentiating both sides w.r.t. x, we get
\(\cos (x+y) \frac{d}{d x}(x+y)=2 x+2 y \frac{d y}{d x} \)
\(\Rightarrow \cos (x+y)\left(1+\frac{d y}{d x}\right)=2 x+2 y \frac{d y}{d x} \)
\(\Rightarrow \cos (x+y)+\cos (x+y) \frac{d y}{d x}=2 x+2 y \frac{d y}{d x}\) 
\(\Rightarrow \frac{d y}{d x}[\cos (x+y)-2 y]=2 x-\cos (x+y) \)
\(\therefore \ \frac{d y}{d x}=\frac{2 x-\cos (x+y)}{\cos (x+y)-2 y}\)

We have,\(\sin ^{2} y+\cos x y=K\)
On differentiating both sides w.r.t. x, we get
\( \frac{d}{d x}\left(\sin ^{2} y+\cos x y\right) \triangleq \frac{d}{d x}(K) \)
\(\Rightarrow \frac{d}{d x}\left(\sin ^{2} y\right)+\frac{d}{d x}(\cos x y) =0 \) 
\( \Rightarrow \frac{d}{d x}\left(\sin ^{2} y\right)+\frac{d}{d x}(\cos x y)=0\)
\(\Rightarrow 2 \sin y \cos y \frac{d y}{d x}+(-\sin x y) \frac{d}{d x}(x y)=0 \)
\(\Rightarrow \sin 2 y \frac{d y}{d x}-\sin x y\left(x \frac{d y}{d x}+y \cdot 1\right)=0 \)
\( \Rightarrow \sin 2 y \frac{d y}{d x}-x \sin x y \frac{d y}{d x}=y \sin x y \)
\( \Rightarrow \frac{d y}{d x}=\frac{y \sin (x y)}{\sin 2 y-x \sin (x y)} \) 
Now, \(\text { at } x=1, y=\frac{n}{4}\) 
\( \therefore\left(\frac{d y}{d x}\right)_{(1, \pi / 4)} =\frac{\frac{\pi}{4} \sin \left(1 \cdot \frac{\pi}{4}\right)}{\sin \left(2 \cdot \frac{\pi}{4}\right)-1 \sin \left(1 \cdot \frac{\pi}{4}\right)} \)
\(=\frac{\frac{\pi}{4} \sin \left(\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{2}\right)-\sin \left(\frac{\pi}{4}\right)}=\frac{\frac{\pi}{4}\left(\frac{1}{\sqrt{2}}\right)}{1-\frac{1}{\sqrt{2}}} \)
\(=\frac{\frac{\pi}{4 \sqrt{2}}}{(\sqrt{2}-1) / \sqrt{2}} \)
\(=\frac{\pi}{4 \sqrt{2}} \times \frac{\sqrt{2}}{(\sqrt{2}-1)}=\frac{\pi}{4(\sqrt{2}-1)} \)

Let \(y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) 
\(\text { Put } x=\tan \theta \Rightarrow \theta=\tan ^{-1} x \)
\( \therefore \ y=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \) 
\(=\cos ^{-1}(\cos 2 \theta)=2 \theta \quad\left[\because \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]\) 
\(\Rightarrow y=2 \tan ^{-1} x\) 
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=2 \times \frac{1}{1+x^{2}}=\frac{2}{1+x^{2}}\) 

Let \(y=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\) 
\(=\tan ^{-1}\left(\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)\) 
\(\left[\because 1+\cos A=2 \cos ^{2} \frac{A}{2} \text { and } \sin A=2 \sin \frac{A}{2} \cos \frac{A}{2}\right]\) 
\(=\tan ^{-1}\left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right)=\tan ^{-1}\left(\cot \frac{x}{2}\right) \)
\(=\frac{\pi}{2}-\cot ^{-1}\left(\cot \frac{x}{2}\right)\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \) 
\(=\frac{\pi}{2}-\frac{x}{2} \quad\left[\because \cot ^{-1}(\cot x)=x\right]\) 
Now, on differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=0-\frac{1}{2}=-\frac{1}{2}\)

We have, \(f(x)=x \cos x+e^{x}\) 
On differentiating both sides w.r.t. x, we get
\( f^{\prime}(x) =\frac{d}{d x}(x \cos x)+\frac{d}{d x}\left(e^{x}\right) \)
\(\Rightarrow f^{\prime}(x) =x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(x)+\frac{d}{d x}\left(e^{x}\right) \)
\(=-x \sin x+\cos x+e^{x} \) 
\(\text { At } x=0 \)
\( f^{\prime}(0)=-(0) \sin 0+\cos 0+e^{0}=0+1+1=2 \)

\((\sin x)^{\log x}\left(\frac{\log \sin x}{x}+\cot x \cdot \log x\right)\)

Use product rule of derivative
\(x e^{x} \cos x+(x+1) e^{x} \sin x\)

Use chain rule of derivative
\(\frac{5}{x \log \left(x^{5}\right) \log \log \left(x^{5}\right)}\)

Given, \(y=\cos (\sqrt{3 x})\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x} =\frac{d}{d x}\{\cos (\sqrt{3 x})\}=-\sin (\sqrt{3 x}) \cdot \frac{d}{d x}(\sqrt{3 x}) \)
\(=-\sin (\sqrt{3 x}) \cdot \frac{1}{2 \sqrt{3 x}} \cdot 3=-\frac{3 \sin \sqrt{3 x}}{2 \sqrt{3 x}}\)

We have, \(\left(x^2+y^2\right)^2=x y\)
On differentiating both sides w.r.t. x, we get
\(2\left(x^2+y^2\right)\left[2 x+2 y \frac{d y}{d x}\right]=\left[x \frac{d y}{d x}+y\right]\)
\(\Rightarrow 4\left(x^2+y^2\right)\left(x+y \frac{d y}{d x}\right)=\left(y+x \frac{d y}{d x}\right)\)
\(\Rightarrow 4\left(x^2+y^2\right) x+4\left(x^2+y^2\right) y \frac{d y}{d x}=y+x \frac{d y}{d x}\)
\(\Rightarrow \frac{d y}{d x}\left[4\left(x^2+y^2\right) y-x\right]=y-4 x\left(x^2+y^2\right) \)
\(\Rightarrow \frac{d y}{d x}=\frac{y-4 x\left(x^2+y^2\right)}{4\left(x^2+y^2\right) y-x} \)
\(\Rightarrow \frac{d y}{d x}=-\frac{\left[y-4 x\left(x^2+y^2\right)\right]}{\left[x-4 y\left(x^2+y^2\right)\right]}\)

Given, \(y=\left(x+\sqrt{x^2-1}\right)^2\)
On differentiating both sides w.r.t. x, we get
\( \frac{d y}{d x}=2\left(x+\sqrt{x^2-1}\right) \frac{d}{d x}\left(x+\sqrt{x^2-1}\right) \)
\(\Rightarrow \frac{d y}{d x}=2\left(x+\sqrt{x^2-1}\right)\left(1+\frac{2 x}{2 \sqrt{x^2-1}}\right) \)
\(=2\left(x+\sqrt{x^2-1}\right)\left(\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\right)\)
\(=\frac{2\left(x+\sqrt{x^2-1}\right)^2}{\sqrt{x^2-1}}=\frac{2 y}{\sqrt{x^2-1}} \)
\(\Rightarrow \quad\left[\because y=\left(x+\sqrt{x^2-1}\right)^2\right]\)
\(\Rightarrow \quad\left(x^2-1\right)\left(\frac{d y}{d x}\right)^2=4 y^2\)
Hence proved.

Given, \(y=x^{1 / x}\)
Taking log on both sides, we get
\(\log y=\frac{1}{x} \log x\)
On differentiating both sides, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(\frac{1}{x}\right)\)
\(\Rightarrow \quad \frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cdot \frac{1}{x}+\log x\left(\frac{-1}{x^2}\right) \)
\(\Rightarrow \quad \frac{d y}{d x}=y\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right)\)
\(\Rightarrow \quad \frac{d y}{d x}=x^{1 / x}\left(\frac{1-\log x}{x^2}\right)\)
\( \therefore \text { At } x=1\)
\(\frac{d y}{d x}=1 \frac{(1-\log 1)}{1}=1-0=1\)

Given, \(x=a \cos t, y=b \sin t\)
On differentiating w.r.t. t, we get
\( \frac{d x}{d t} =a(-\sin t)=-a \sin t \)
\(\text { and }  \frac{d y}{d t}  =b \cos t \)
\(\therefore  \frac{d y}{d x} =\frac{d y / d t}{d x / d t}=\frac{b \cos t}{-a \sin t}=\frac{-b}{a} \cot t\)
Again, differentiating both sides w,r.t. x, we get 
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{-b}{a} \cot t\right)=\frac{-b}{a} \frac{d}{d t}(\cot t) \cdot \frac{d t}{d x}\)
\(=\frac{-b}{a}\left(-\operatorname{cosec}^2 t\right)\left(\frac{-1}{a \sin t}\right)=\frac{-b}{a^2} \operatorname{cosec}^3 t\)

Given, \(y=\sqrt{a x+b}\)
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{a x+b}} \frac{d}{d x}(a x+b) \)
\(\Rightarrow  \frac{d y}{d x}=\frac{a}{2 \sqrt{a x+b}}=\frac{a}{2 y} \)
\(\Rightarrow  y \frac{d y}{d x}=\frac{a}{2}\)
Again, differentiating both sides, w.r.t. x, we get
\(y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{d y}{d x}=0\)
\(\Rightarrow  y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2=0\)
Hence proved.

We have, \(y=e^{x \sin ^2 x}+(\sin x)^x\)
Let \(u=e^{x \sin ^2 x}\) and \(v=(\sin x)^x\)
Now, \(u=e^{x \sin ^2 x}\)
\(\Rightarrow \quad \frac{d u}{d x}  =e^{x \sin ^2 x} \quad\left[1 \cdot \sin ^2 x+x(2 \sin x \cos x)\right] \)
\(=e^{x \sin ^2 x}\left[\sin ^2 x+x \sin 2 x\right]\)
\( \therefore \quad y=u+v\)
\( \Rightarrow \quad \frac{d y}{d x}=\frac{d t}{d x}+\frac{d v}{d x}\)
\(\text { Also, }  y=(\sin x)^x\)
\(\Rightarrow  \log v=\log (\sin x)^x=x \log \sin x\)
\(\Rightarrow \quad \frac{1}{v} \frac{d v}{d x}=1 \cdot \log \sin x+x \cdot\left(\frac{1}{\sin x}\right) \cos x\)
\(\Rightarrow \quad \frac{d v}{d x}=v[\log \sin x+x \cot x]\)
\(\Rightarrow \frac{d v}{d x}=(\sin x)^x[\log \sin x+x \cot x]\)
\(\therefore \text { From Eqs. (i), (ii) and (iii), we have }\)
\(\frac{d y}{d x}=e^{x \sin ^2 x}\left[\sin ^2 x\right. +x \sin 2 x] +(\sin x)^x[\log \sin x+x \cot x](1)\)