The given function is \(f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}=\frac{g(x)}{h(x)}\) where g (x) = sin (ax + b) and h (x) = cos (cx + d)
\(\therefore f^{\prime}=\frac{g^{\prime} h-g h^{\prime}}{h^{2}} \)
\(\text { Consider } g(x)=\sin (a x+b) \)
\(\text { Let } u(x)=a x+b, v(t)=\sin t \)
\(\text { Then, }(\text { vou })(x)=v(u(x))=v(a x+b)=\sin (a x+b)=g(x) \)
∴ g is a composite function of two functions, u and v.
\(\text { Put } t=u(x)=a x+b \)
\(\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b) \)
\(\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a \)
Therefore, by chain rule, we obtain
\(g^{\prime}=\frac{d g}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b) \)
\(\text { Consider } h(x)=\cos (c x+d) \)
\(\text { Let } p(x)=c x+d, q(y)=\cos y \)
\(\text { Then, }(q o p)(x)=q(p(x))=q(c x+d)=\cos (c x+d)=h(x) \)
∴h is a composite function of two functions, p and q.
Put y = p (x) = cx + d
\(\frac{d q}{d y}=\frac{d}{d y}(\cos y)=-\sin y=-\sin (c x+d) \)
\(\frac{d y}{d x}=\frac{d}{d x}(c x+d)=\frac{d}{d x}(c x)+\frac{d}{d x}(d)=c \)
Therefore, by chain rule, we obtain
\(h^{\prime}=\frac{d h}{d x}=\frac{d q}{d y} \cdot \frac{d y}{d x}=-\sin (c x+d) \times c=-c \sin (c x+d)\)
\(\therefore f^{\prime}=\frac{a \cos (a x+b) \cdot \cos (c x+d)-\sin (a x+b)\{-c \sin (c x+d)\}}{[\cos (c x+d)]^{2}} \)
\(=\frac{a \cos (a x+b)}{\cos (c x+d)}+c \sin (a x+b) \cdot \frac{\sin (c x+d)}{\cos (c x+d)} \times \frac{1}{\cos (c x+d)} \)
\(=a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) \tan (c x+d) \sec (c x+d)\)

Here g(x) = x - [x]
Let a be an integer and h is very small h > 0, then
\( [a-h]=a-1,[a+h]=a \)
\(\text {and } [a]=a \) 
\(\text {At } x=a, \mathrm{LHL}=\lim _{x \rightarrow a^{-}} g(x)=\lim _{x \rightarrow a^{-}}(x-[x])\)
\(Put x=a-h ; when x \rightarrow a^{-}, then\ h \rightarrow 0\)
\( \mathrm{LHL} =\lim _{h \rightarrow 0}(a-h-[a-h]) \)
\(=\lim _{h \rightarrow 0}(a-h-(a-1)) \)

\(\Rightarrow \frac { dy }{ dx } ={ x }^{ { x }^{ x } }.{ x }^{ x }.\ log\ x\left[ 1+log \ \ x\ +\frac { 1 }{ x\quad log\quad x } \right] \)

\(\Rightarrow y=\frac { x }{ 2 } \Rightarrow y'=\frac { 1 }{ 2 } \)

\(\frac { dy }{ dx } ={ e }^{ x }\left[ 2cot2x+log\quad (sin2x) \right] \)

\({ y }^{ ' }=\frac { 2 }{ 1+x^{ 2 } } \)

\(\Rightarrow \) \(\left( { 1-x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\frac { { d }y }{ d{ x } } =0\)\(\left[ On\quad dividing\quad by\quad 2\left( \frac { dy }{ dx } \right) \right] \)

\(\frac { dy }{ dx } ={ 6e }^{ 2x }+{ { 6e }^{ 3x } },\frac { { d }^{ 2 }y }{ dx^{ 2 } } =12{ e }^{ 2x }+18{ e }^{ 3x }\) 
substitute in L.H.S.

\(\frac { dy }{ dx } =\frac { -1 }{ { (1+x })^{ 2 } } \)

\(\Rightarrow { x }^{ 2 }{ y }^{ ' }+xy^{ ' }+y=0\)

\(\text {Let } y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}\)
\(\text {Also, let } u=(\sin x)^{x} \text { and } v=\sin ^{-1} \sqrt{x}\)
\(\therefore y=u+v\)
\(\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(u=(\sin x)^{x}\)
\(\Rightarrow \log u=\log (\sin x)^{x}\)
\(\Rightarrow \log u=x \log (\sin x)\)
Differentiating both sides with respect to x  we obtain
\(\Rightarrow \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \times \frac{d}{d x}[\log (\sin x)]\)
\(\Rightarrow \frac{d u}{d x}=u\left[1 \cdot \log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]\)
\(\Rightarrow \frac{d u}{d x}=(\sin x)^{x}\left[\log (\sin x)+\frac{x}{\sin x} \cdot \cos x\right] \)
\(\Rightarrow \frac{d u}{d x}=(\sin x)^{x}(x \cot x+\log \sin x) \)
\(v=\sin ^{-1} \sqrt{x} \)
Differentiating both sides with respect to x, we obtain
\(\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \)
\(\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \)
\(\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}} \)
Therefore, from (1), (2), and (3), we obtain
\(\frac{d y}{d x}=(\sin x)^{x}(x \cot x+\log \sin x)+\frac{1}{2 \sqrt{x-x^{2}}}\)

As in the example we find that f is continuous at all real numbers x \(\ne\) 1. The left hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+2)=1+2=3\)
The right hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-2)=1-2=-1\)
Since, the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f.

For continuity at x = 0, we have
\(\underset{x=0}{\mathrm{LHL}}=\underset{x=0}{\mathrm{RHL}}=f(0)\)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^{2}}=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}=a \)
\(\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^{2} 2 x}{x^{2}}=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}})+4}{16+\sqrt{x}-16}=a \)
\(\Rightarrow \lim _{x \rightarrow 0} 8\left(\frac{\sin 2 x}{2 x}\right)^{2}=\lim _{x \rightarrow 0^{+}}(\sqrt{16+\sqrt{x}}+4)=a \)
\(\Rightarrow 8 \times(1)^{2}=4+4=a \Rightarrow a=8 \)
a = 8

K=-1, for function to be continuous at x=0

Here, \(f(x)=\left\{\begin{array}{cl} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.\) 
At \(x=0, \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)\) 
\(\therefore \mathrm{LHL}=\lim _{h \rightarrow 0} \lambda\left[(0-h)^{2}-2(0-h)\right]=\lim _{h \rightarrow 0}\left[\lambda\left(h^{2}+2 h\right)\right]=0\) 
\(\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)\) 
\(\therefore \mathrm{RHL}=\lim _{h \rightarrow 0}[4(0+h)+1]=\lim _{h \rightarrow 0}[4 h+1]=0+1=1\)
\(\text { [put } x=0+h \text { ; when } x \rightarrow 0^{+} \text {, then } \left.h \rightarrow 0\right] \) 
\(\therefore \mathrm{LHL} \neq \mathrm{RHL}\) 
Thus, f(x) is not continuous at x = 0 for any value of λ.
At x = 1,
 \( \mathrm{LHL} =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(4 x+1) \)
\(\therefore \mathrm{LHL} =\lim _{h \rightarrow 0}[4(1-h)+1]=\lim _{h \rightarrow 0}[5-4 h]=5-0=5 \)
[put x=1−h; when x→1−,then h→0]   
\( \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x+1) \)
\(\therefore \lim _{h \rightarrow 0}[4(1+h)+1]=\lim _{h \rightarrow 0}(5+4 h)=5+0=5 \)
[put x=1+h; when x→1, then h→0]   
Also, f(1)=4×1+1=5
\([\because f(x)=4 x+1]\)
Thus f(x) is continuous at x=1 for all values of λ.