\( \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}\right.\left.-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right] \)

(a) we have:
f(x) = sin + cos x,
\({ D }_{ f }=R\)
\(Let\ c\in { D }_{ f }\)
\(Now\ \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ (sin+cosx) } \)
f is continuous function.
(b) Replace (+) by (-)
(c) Replace (+) by (.)

\(f(x)=\left\{\begin{array}{rll} \frac{k \cos x}{\pi-2 x}, & \text { if } & x \neq \frac{\pi}{2} \\ 3, & \text { if } & x=\frac{\pi}{2} \end{array}\right.\)
f(x) is continuous at \(x=\pi / 2\)
So
\( \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=f\left(\frac{\pi}{2}\right) \)
\(​​​​​​​\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi^*}{2}} \frac{k \cos x}{\pi-2 x}=3 \)
\( \lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}=3 \)
 \(\lim _{h \rightarrow 0} \frac{k \sin h}{2 h}=\lim _{h \rightarrow 0} \frac{-k \sin h}{-2 h}=3\)
\( \frac{k}{2}(1)=3 \Rightarrow k=6\)

\(We\quad have:\quad x={ at }^{ 2 },\quad y=2at.\)
\(\frac { dx }{ dt } =2at\quad and\quad \frac { dy }{ dt } =2a.\)
\(\frac { dy }{ dx } =\frac { { dy }/{ dt } }{ { dx }/{ dt } } =\frac { 2a }{ 2at } =\frac { 1 }{ t } ,t\neq 0\)

\(x^y+y^x=1\)
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 )

Observe that the given function is a composite of two functions. Indeed, if t = u(x) = x² and v(t) = sin t, then
f(x) = (v o u) (x) = v(u(x)) = v(x 2 ) = sin x²
\(\text {Put } t=u(x)=x^2 \text {. Observe that } \frac{d v}{d t}=\cos t \text { and } \frac{d t}{d x}=2 x\) exist. Hence, by chain rule
\(\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos t \cdot 2 x\)
It is normal practice to express the final result only in terms of x. Thus
\(\frac{d f}{d x}=\cos t \cdot 2 x=2 x \cos x^2\)

\(We\quad have:f(x)={ x }^{ 3 }+{ x }^{ 2 }-1\)
Which is polynomial function
\(and\ { D }_{ f }=R\)
\(Let\quad c\in { D }_{ f }\)
\(Then\ \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ ({ x }^{ 3 }+{ x }^{ 2 }-1) } \)
\(={ c }^{ 3 }+{ c }^{ 2 }-1=f(c)\)
\(\Rightarrow \) f is continuous at x = c.
But c is arbitrary.
Hence, f is continuous at each of its domains.

The function is defined at x = 0 and its value at x = 0 is 1. When x \(\ne\) 0, the function is given by a polynomial. Hence
\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ { x }^{ 3 }+3 } =0+3=3\)
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function.

First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly
\(\lim _{ x\rightarrow 1 }{ f(x) } =\lim _{ x\rightarrow 1 }{ \left( { 2x }+3 \right) } =2(1)+3=5\)
\(f(1)=2(1)+3=5.\)
\(Thus\quad \lim _{ x\rightarrow 1 }{ f(x) } =f(1)\)
Hence,f is continuous at x = 1.

\(\text {(i) Let } y=\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^2+4}}=(3 x+2)^{\frac{1}{2}}+\left(2 x^2+4\right)^{-\frac{1}{2}}\)
Note that this function is defined at all real numbers \(x>-\frac{2}{3}\)Therefore
\( \frac{d y}{d x} =\frac{1}{2}(3 x+2)^{\frac{1}{2}-1} \cdot \frac{d}{d x}(3 x+2)+\left(-\frac{1}{2}\right)\left(2 x^2+4\right)^{-\frac{1}{2}-1} \cdot \frac{d}{d x}\left(2 x^2+4\right) \\ \)
\(=\frac{1}{2}(3 x+2)^{-\frac{1}{2}} \cdot(3)-\left(\frac{1}{2}\right)\left(2 x^2+4\right)^{-\frac{3}{2}} \cdot 4 x \\ \)
\( =\frac{3}{2 \sqrt{3 x+2}}-\frac{2 x}{\left(2 x^2+4\right)^{\frac{3}{2}}}\)
This is defined for all real numbers \(x>-\frac{2}{3}\)
\(\text {(ii) Let } y=\log _7(\log x)=\frac{\log (\log x)}{\log 7}\) (by change of base formula).
The function is defined for all real numbers x > 1. Therefore
\( \frac{d y}{d x} =\frac{1}{\log 7} \frac{d}{d x}(\log (\log x)) \)
\( =\frac{1}{\log 7} \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \)
\( =\frac{1}{x \log 7 \log x}\)

The function y = (sin x)^{sin x} is defined for all positive real numbers. Taking logarithms, we have
log y = log (sin x)^{sin x} = sin x log (sin x)
Then \(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(\sin x \log (\sin x))\)
\( =\cos x \log (\sin x)+\sin x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x) \)
\(=\cos x \log (\sin x)+\cos x \)
\(=(1+\log (\sin x)) \cos x\)
Thus, \(\frac{d y}{d x}=y((1+\log (\sin x)) \cos x)=(1+\log (\sin x))(\sin x)^{\sin x} \cos x\)

Let \(y=(\sin x)^x+\sin ^{-1} \sqrt{x}\)
Let \(u=(\sin x)^x \ v=\sin ^{-1} \sqrt{x}\)
y=u+v
Differentiating both sides w.r.t. x.
\( \frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)

Chain rule may be extended as follows. Suppose f is a real valued function which is a composite of three functions u, v and w; i.e.,
f = (w o u) o v. If t = v (x) and s = u (t), then
\(\frac{d f}{d x}=\frac{d(w \circ u)}{d t} \cdot \frac{d t}{d x}=\frac{d w \mid}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x}\)
provided all the derivatives in the statement exist. Reader is invited to formulate chain rule for composite of more functions.

\(x^{x^2-3}+(x-3)^{x^2}, \text { for } x>3\)
Let \(y=x^{x^2-3}+(x-3)^{x^2}\)
And let \(u=x^{x^2-3}, v=(x-3)^{x^2}\)
\(\boldsymbol{y}=\boldsymbol{u}+\boldsymbol{v}\)
Differentiating both sides w.r.t. x
\(\frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Differentiating both sides with respect to x, we obtain
\( \frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^2-3\right)+\left(x^2-3\right) \cdot \frac{d}{d x}(\log x) \)
\( \Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^2-3\right) \cdot \frac{1}{x} \)
\( \Rightarrow \frac{d u}{d x}=x^{x^2-1}\left[\frac{x^2-3}{x}+2 x \log x\right]\)
Now, v=(x-3)^x
Taking logarithm on both the sides, we obtain
\(\log v =\log (x-3)^{x^2} =x^2 \log (x-3)\)
Differentiating both sides with respect to x, we obtain
\(\frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^2\right)+\left(x^2\right) \cdot \frac{d}{d x}[\log (x-3)] \)
\(\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^2 \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \)
\( \Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^2}{x-3} \cdot 1\right] \)
\( \Rightarrow \frac{d v}{d x}=(x-3)^{x^2}\left[\frac{x^2}{x-3}+2 x \log (x-3)\right]\)
From (1), (2), and (3), we obtain

We are investigating continuity of (f + g) at x = c. Clearly it is defined at
x = c. We have
\( \lim _{x \rightarrow c}(f+g)(x) =\lim _{x \rightarrow c}[f(x)+g(x)] \\ =\lim _{x \rightarrow c} f(x)+\lim _{x \rightarrow c} g(x) \\ =f(c)+g(c) \\ =(f+g)(c) \)
Hence, f + g is continuous at x = c.
Proofs for the remaining parts are similar and left as an exercise to the reader.