consider the function f(x) = |x - 1| + |x - 2|
Differentiability of f(x)
Since modulus function is everywhere continuous and sum of two continuous function is also continuous.Graph of f(x) shows, that f(x) is every where derivable except possible at x = 0 and x = 1

It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.

The given function is cos x³ . sin² (x⁵)
\(\frac{d}{d x}\left[\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)\right]=\sin ^{2}\left(x^{5}\right) \times \frac{d}{d x}\left(\cos x^{3}\right)+\cos x^{3} \times \frac{d}{d x}\left[\sin ^{2}\left(x^{5}\right)\right] \)
\(=\sin ^{2}\left(x^{5}\right) \times\left(-\sin x^{3}\right) \times \frac{d}{d x}\left(x^{3}\right)+\cos x^{3} \times 2 \sin \left(x^{3}\right) \cdot \frac{d}{d x}\left[\sin x^{5}\right] \)
\(=-\sin x^{3} \sin ^{2}\left(x^{5}\right) \times 3 x^{2}+2 \sin x^{5} \cos x^{3} \cdot \cos x^{5} \times \frac{d}{d x}\left(x^{5}\right) \)
\(=-3 x^{2} \sin x^{3} \cdot \sin ^{2}\left(x^{5}\right)+2 \sin x^{5} \cos x^{5} \cos x^{3} \cdot \times 5 x^{4} \)
\(=10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2}\left(x^{5}\right) \)

\(f(x)=\left\{\begin{array}{ll} x+1, & \text { if } x \geq 1 \\ x^{2}+1, & \text { if } x<1 \end{array}\right.\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:\(\text { If } c<1, \text { then } f(c)=c^{2}+1 \text { and } \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}+1\right)=c^{2}+1\)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x < 1
Case II: \(\text { If } c=1, \text { then } f(c)=f(1)=1+1=2\)
The left hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{2}+1\right)=1^{2}+1=2\)
The right hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2 \)
\(\therefore \lim _{x \rightarrow 1} f(x)=f(1) \)
Therefore, f is continuous at x = 1
Case III: \(\text { If } c>1, \text { then } f(c)=c+1 \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1 \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x > 1
Hence, the given function f has no point of discontinuity

\(f(x)=\left\{\begin{array}{ll} \frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0 \end{array}\right.\)
It is known that \(x<0 \Rightarrow|x|=-x\)
Therefore, the given function can be rewritten as
\(f(x)=\left\{\begin{array}{l} \frac{x}{|x|}=\frac{x}{-x}=-1, \text { if } x<0 \\ -1, \text { if } x \geq 0 \end{array}\right.\)
\(\Rightarrow f(x)=-1 \text { for all } x \in \mathbf{R} \)
Let c be any real number. Then, \(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1\)
Also \(f(c)=-1=\lim _{x \rightarrow c} f(x)\)
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.

If function is continuous at \(x=\frac{\pi}{2}, \text { then }\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\mathrm{RHL}=f\left(\frac{\pi}{2}\right)\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \)
\(=\lim _{h \rightarrow 0} \frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}=\lim _{h-0} \frac{1-\cos ^{3} h}{3 \sin ^{2} h} \)
\(=\lim _{h \rightarrow 0} \frac{(1-\cos h)\left(1+\cos ^{2} h+\cos h\right)}{3(1-\cos h)(1+\cos h)} \)
\(=\lim _{h \rightarrow 0} \frac{1+\cos ^{2} h+\cos h}{3(1+\cos h)}=\frac{1+1+1}{3(1+1)}=\frac{1}{2} \ldots .(i i) \)
\(\operatorname{RHL} =\lim _{h=\frac{\pi}{2}} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0} \frac{b\left\{1-\sin \left(\frac{\pi}{2}+h\right)\right\}}{\left\{\pi-2\left(\frac{\pi}{2}+h\right)\right\}^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{(\pi-\pi-2 h)^{2}}=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{4 h^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b \cdot 2 \sin ^{2} \frac{h}{2}}{4 h^{2}}=\lim _{h \rightarrow 0} \frac{b}{8}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \)
\(=\frac{b}{8} \times 1=\frac{b}{8} \)
Substituting from (ii) and (iii) in (i), we get
\(\frac{1}{2}=\frac{b}{8}=a \Rightarrow a=\frac{1}{2}, b=4\)
Hence, for \(a=\frac { 1 }{ 2 } ,\quad b=4 \text { function is continuous at} x=\frac { \pi }{ 2 } .\)

\(Let\ y={ \left( 5x \right) }^{ 3\ cos\ 2x }\)
\(Taking\ logs.,\ log\ y=log\ { \left( 5x \right) }^{ 3\ cos\ 2x }\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ y } .\frac { dy }{ dx } \)
\(=3\left[ cos\quad 2x.\frac { 1 }{ 5x } (5)+log(5x).(-sin\quad 2x)2 \right] \)
\(\Rightarrow \frac { dy }{ dx } \)
\(=3y\left[ \frac { cos\quad 2x }{ x } -2sin\quad 2x\quad log\quad 5x \right]\)
\( Hence\quad \frac { dy }{ dx } ={ \left( 5x \right) }^{ 3\quad cos\quad 2x }\left[ 3\frac { cos\quad 2x }{ x } -6sin\quad 2x\quad log\quad 5x \right] \)

Observe that both y and x are defined for all real t ≠ 0. Clearly
\( \frac{d y}{d t}=\frac{d}{d t}\left(a^{t+\frac{1}{t}}\right) =a^{t+\frac{1}{t}} \frac{d}{d t}\left(t+\frac{1}{t}\right) \cdot \log a \\ =a^{t+\frac{1}{t}}\left(1-\frac{1}{t^2}\right) \log a \)
Similarly \(\frac{d x}{d t} =a\left[t+\frac{1}{t}\right]^{a-1} \cdot \frac{d}{d t}\left(t+\frac{1}{t}\right) \)
\( =a\left[t+\frac{1}{t}\right]^{a-1} \cdot\left(1-\frac{1}{t^2}\right) \)
\(\frac{d x}{d t} \neq 0 \text { only if } t \neq \pm 1 \text {. Thus for } t \neq \pm 1\)
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a^{t+\frac{1}{t}}\left(1-\frac{1}{t^2}\right) \log a}{a\left[t+\frac{1}{t}\right]^{a-1} \cdot\left(1-\frac{1}{t^2}\right)}\)
\(=\frac{a^{t+\frac{1}{t}} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}\)

\(We\quad have:\quad { e }^{ y }(x+1)=1\)
\({ e }^{ y }=\frac { 1 }{ x+1 } \)
\(y=log\left( \frac { 1 }{ x+1 } \right) \)
\(y=-log(x+1)......(1)\)
\(\frac { dy }{ dx } =-\frac { 1 }{ x+1 } ....(2)\)
\( and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(Hence,\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)

\(We\quad have\quad y=5\quad cosx-3sinx\)
\( \frac { dy }{ dx } =-5\quad sin\quad x-3\quad cos\quad x\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-5\quad cos\quad x+3\quad sinx\)
\(=-(5\quad cos\quad x-3\quad sin\quad x)\)
\(=-y\)
\(Hence,\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

\(Let\quad y={ x }^{ 2 }+3x+2\)
\(\frac { dy }{ dx } =2x+3\quad and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\)

\(Let\ y=cosx.\ cos\ 2x.\ cos\ 3x\)
\(Taking\ logs,\)
\(log\ y=log(os\ x.os\ 2x.\ cos\ 3x)\)
\(=log\ cosx+log\ cos2x+\ log\ cos3x\)
\(Diff.\quad w.r.t.\quad \frac { 1 }{ y } \frac { dy }{ dx } \)
\(=\frac { 1 }{ cos\quad x } (-sin\quad x)+\frac { 1 }{ cos\quad 2x } (-2sin\quad 2x)+\frac { 1 }{ cos\quad 3x } (-sin\quad 3x)\)
\(\frac { dy }{ dx } =y\left[ -tan\quad x-2\quad tan\quad 2x-3\quad tan\quad 3x \right] \)
\(=cos\quad x.os\quad 2x.\quad cos\quad 3x\)
\(\left[ -tan\ x-2\ tan\ 2x-3\ tan3x \right] \)
\(=-cos\ x\ cos\ 2x\ cos\ 3x\)

\(We\ have:\ f(x)={ x }^{ 2 }-sin\ \pi +5={ \pi }^{ 2 }-0+5\)
\(={ \pi }^{ 2 }+5.\)
\(Now\ \lim _{ x\rightarrow \pi }{ f(x) } =\lim _{ x\rightarrow \pi }{ f({ x }^{ 2 }-sinx+5) } \)
\(={ \pi }^{ 2 }-sin\pi +5.\)
\(={ \pi }^{ 2 }-0+5={ \pi }^{ 2 }+5=f(\pi )\)
\(Hence,\ 'f'\ is\ continuous\ at\ x=\pi \)

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.

\(\text {(i) Let } y=\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^2+4}}=(3 x+2)^{\frac{1}{2}}+\left(2 x^2+4\right)^{-\frac{1}{2}}\)
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
\( \frac{d y}{d x} =\frac{1}{\log 7} \frac{d}{d x}(\log (\log x)) \)
\( =\frac{1}{\log 7} \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \)
\( =\frac{1}{x \log 7 \log x}\)

\(x^{x^2-3}+(x-3)^{x^2}, \text { for } x>3\)
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 have,
\(f(x)=x^{2}+2 x-8, x \in[-4,2]\) 
Rolle'stheoremis satisfied
\( \therefore f^{\prime}(c)=0 \)
\(\Rightarrow f^{\prime}(c)=2 c+2=0 \)
\(c+1=0 \Rightarrow c=-1 \) 

Let y = a^x_.
Taking log on both sides, we get log y = x log a
On differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\log a \Rightarrow \frac{d y}{d x}=y \log a\) 
Thus,\(\frac{d}{d x}\left(a^{x}\right)=a^{x} \log a\)

Given, y = tan^-1\(\sqrt { \frac { sinx }{ 1+cosx } } \)
y = tan^-1\(\sqrt { \frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } } \)
y = tan^-1\(\left( \sqrt { tan\frac { x }{ 2 } } \right) \)
y = tan^-1

\(\frac { d }{ dx } \left[ sin\left\{ log\left( { x }^{ 2 }-1 \right) \right\} \right] \)
\(=cos\left[ log\left( { x }^{ 2 }-1 \right) \right] .\frac { 1 }{ { x }^{ 3 }-1 } .3{ x }^{ 2 }\)
\(=\frac { 3{ x }^{ 2 }cos\left[ log\left( { x }^{ 3 }-1 \right) \right] }{ { x }^{ 3 }-1 } \)

f(x) = [x]  is not continuous for integers.Hence not continuous at x = ±2, ±1, 0