Step (i) f(x₀) = 7x₀ + 5.
For any \(\triangle x\neq 0,\),
f(x₀ +\(\triangle\)x) = 7(x₀ + \(\triangle\)x) + 5
= 7x₀ + 7\(\triangle\)x + 5
Step (ii) \(\triangle\)y = f(x₀ + \(\triangle\)x) - f(x₀)
= (7x_o+7\(\triangle\)x_o+5)
Step (iii)\({\triangle \ y\over \triangle x}=7\)
Thus, at any point on the graph of f(x) = 7x + 5, we have
Step (iv) m_tan = \(lim_{\triangle x \rightarrow 0}{\triangle y\over \triangle x}\)
= \(lim_{\triangle x\rightarrow o}(7)\)
= 7.

The greatest integer function \(f(x)=\lfloor x\rfloor\) is not continuous at every integer point n, since \(\left.\lim _{x \rightarrow n^{-}} \mid x\right\rfloor=n-1\) and \(\lim _{x \rightarrow n^{+}}\lfloor x\rfloor=n .\) Thus f'(n) does not exist.

f'(x) = \(\lim _{h \rightarrow 2} \frac{f(x+h)-f(x)}{h}\)
Given f(x) = 6
⇒ f(x + hx) = 6
∴ f'(x) = \(\lim _{h \rightarrow 0} \frac{6-6}{h}=\lim _{t \rightarrow 0} \frac{0}{h}\) = 0 
∴ f'(x) = 0

\(f(x)=-x^2+2\)
\(f(x+h)=-(x+h)^2+2=-x^2-h^2-2 x h+2\)
\(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+2+x^2-2}{h}\)
\(=\lim _{h \rightarrow 0} \frac{+h(-h-2 x)}{h}\)
= -0 - 2x
\(f^{\prime}(x)=-2 x\)

\(f(x)=\left\{\begin{array}{ccc} 1-2 x & \text { if } & x<0 \\ 1 & \text { if } & 0 \leq x<1 \\ 2 x-1 & \text { if } & x \geq 1 \end{array}\right.\)
At x = 0
\(f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{x \rightarrow 0^{+}} \frac{1-1}{x-0}=0\)
\(f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{x \rightarrow 0^{-}} \frac{1-2 x-1}{x-0}=-\infty\)
\(\therefore\) It is not differentiable at x = 0
At x = 1
\(f^{\prime}\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{2(x-1)}{(x-1)}=2\)
\(f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{1-1}{x-1}=0\)
\(\therefore f^{\prime}\left(1^{+}\right) \neq f^{\prime}\left(1^{-}\right)\)
\(\therefore f\) is not differentiable at x = 1.
Hence f(x) i not differentiable at x = 0, 1.

y = sinx

\(y=|\sin x|\)

\(|\sin x|\) is not differentiable at \(x=n \pi, n \in Z\)

Let y = 2^x = e^xlog2.
Take u = (log 2)x so that
y = e^u
\({dy\over dx}={dy\over du}\times {du\over dx}=e^u \times log2=e^{xlog 2}\)
= (log2)2^x.

y = cos(tan x)
Take \(u=\tan x \Rightarrow \frac{d u}{d x}=\sec ^2 x\)
\(y=\cos u\)
\(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=-\sin u \cdot\left(\sec ^2 x\right)\)
\(=-\sin (\tan x) \sec ^2 x\)

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

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

\(\therefore\) f'(x) = \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { 3x-0 }{ x } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { 3x }{ x } = 3\) [f(x) = 3x]  .....(1)
\(\therefore\) f'(0⁺) = \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { -4x-0 }{ x } \) 
= \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { -4x }{ x } \) = -4 .........(2)
by (1) & (2), f'(0^-) ≠ f'(0⁺)
\(\therefore\) It is not differentiable.

\(\frac{d y}{d x}=e^x+\cos x\)

\(y={log \ x \over e^x }=e^{-x}.log \ x\)
\({dy\over dx}=e^{-x}({1\over x})+log \ x(e^{-x})(-1)\)
\(=e^{-x}[{1\over x}-log \ x]\).

\(y=\frac{x}{\sin X+\cos X}\)
\(\frac{d y}{d x}=\frac{(\sin x+\cos x) \frac{d}{d x}(x)-x \frac{d}{d x}(\sin x+\cos x)}{(\sin x+\cos x)^2}\)
\(=\frac{(\sin x+\cos x)(1)-x(\cos x-\sin x)}{(\sin x+\cos x)^2}\)
\(=\frac{\sin x+\cos x-x \cos x+x \sin x}{(\sin x+\cos x)^2}\)
\(=\frac{\sin x(1+x)+\cos x(1-x)}{(\sin x+\cos x)^2}\)

Take logarithm :
log y = \(\sqrt{x} \log x\)
Differentiating implicitly,
\({y'\over y}=\sqrt{x}.{1\over x}+{1\over 2\sqrt{x}}.log\ x\)
\(={log \ x+2\over 2\sqrt{x}}\)
Therefore, \({d\over dx}(x^{\sqrt{x}})=y'=x^{{\sqrt{x}}}({log \ x+2\over 2\sqrt{x}})\) .

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\(({-1\over \sqrt{1-x^2}})={-3\over \sqrt{1-x^2}}.\)

\( \sqrt{x^2+y^2} =\tan ^{-1} \frac{y}{x} \)
\(\left(x^2+y^2\right)^{1 / 2} =\tan ^{-1} \frac{y}{x}\)
\( \frac{1}{2 \sqrt{x^2+y^2}}\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{1+\left(\frac{y}{x}\right)^2}\left(\frac{x \frac{a y}{d x}-y}{x^2}\right) \)
\(\frac{2\left(x+y \frac{d y}{d x}\right)}{2 \sqrt{x^2+y^2}}=\frac{1}{\left(\frac{x^2+y^2}{x^2}\right)}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)\)
\( x+y \frac{d y}{d x}=\frac{\sqrt{x^2+y^2}}{\left(x^2+y^2\right)}\left(x \frac{d y}{d x}-y\right) \)
\(x+y \frac{d y}{d x}=\frac{1}{\sqrt{x^2+y^2}}\left(x \frac{d y}{d x}-y\right) \)
\(x+y \frac{d y}{d x}=\frac{x}{\sqrt{x^2+y^2}} \frac{d y}{d x}-\frac{y}{\sqrt{x^2+y^2}}\)
\( \frac{x}{\sqrt{x^2+y^2}} \frac{d y}{d x}-y \frac{d y}{d x} =x+\frac{y}{\sqrt{x^2+y^2}}\)
\(\frac{d y}{d x}\left(\frac{x}{\sqrt{x^2+y^2}}-y\right) =\frac{x \sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}} \)
\(\frac{d y}{d x}\left(\frac{x-y \sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\right) =\frac{x \sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}} \)
\(\frac{d y}{d x} =\frac{x \sqrt{x^2+y^2}+y}{x-y \sqrt{x^2+y^2}}\)

\(u =\frac{2}{3} x^{3}+5 \)
\(\frac{d u}{d x} =\frac{2}{3}\left(3 x^{2}\right)=2 x^{2} \)
\(y =\frac{1}{4} u^{4} \)
\(\frac{d y}{d x} =\frac{1}{4}\left(4 u^{3}\right) \frac{d u}{d x}=u^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2 x^{3}+15}{3}\right)^{3} \times 2 x^{2}=\frac{\left(2 x^{3}+15\right)^{3}}{27}\left(2 x^{2}\right) \)

\(y =\cos \left(\sin x^{2}\right) \)
\(\frac{d y}{d x} =-\sin \left(\sin x^{2}\right) \cos \left(x^{2}\right)(2 x) \)
\(\text { At } x =\sqrt{\pi / 2}, \frac{d y}{d x}=-\sin \sin \left(\frac{\pi}{z}\right) \cos (\pi / 2) 2(\sqrt{\pi / 2}) \)
\(=(\sin 1)(0) 2\left(\frac{\sqrt{\pi}}{2}\right)=0 \quad[\because \cos \pi / 2=0] \)

\(y =m x+c \)
\(f(x) =m x+c \Rightarrow f(0)=c=1 \)
\(\therefore c =1 \)
\(f^{\prime}(x) =m \)
\(f^{\prime}(0) =1=m \)
\(\therefore m =1 \)
\(\therefore f(x) =x+1 \)
\(f(2) =2+1=3 \)

\(\text { Given } f(x)=x \tan ^{-1} x\)
\(f^{\prime}(x) =x \cdot \frac{1}{1+x^{2}}+\tan ^{-1} x \)
\(f^{\prime}(1) =\frac{1}{1+1^{2}}+\tan ^{-1}(1) \)
\(=\frac{1}{2}+\pi / 4 \)
\(\left(\because \tan 45^{\circ}=1\right)\)

\(\text { Given } f(x)=x+2\)
\(f^{\prime}(x) =1 \)
\(f^{\prime}(f(x)) =f^{\prime}(x+2)=1 \)

\(\text { The right hand derivative of } f(x) \text { at } x=2 \text { is }\)
\(f^{\prime}\left(2^{+}\right) =\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{(3 x+4)-(3(2)+4)}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{3 x+4-6-4}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{3(x-2)}{(x-2)}=3 \)

\(\text { Given } f(0)=5\)
\(\lim _{x \rightarrow 0} \frac{f(x)-5}{x} =4 \)
\(\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} =4 \)
\(f^{\prime}(0) =4 \)
\(g(x) =\left(x^{2}+2 x+1\right) f(x) \)
\(g^{\prime}(x) =\left(x^{2}+2 x+1\right) f^{\prime}(x)+(2 x+2) f(x) \)
\(g^{\prime}(0) =(1) f^{\prime}(0)+2 f(0) \)
\(=1(4)+2(5) \)
\(=4+10=14 \)

\(f^{\prime}\left(3^{-}\right)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{-}} \frac{x+2-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{-}} \frac{x-3}{x-3}=1\)
\(f^{\prime}\left(3^{+}\right)=\lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{+}} \frac{8-x-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{+}} \frac{3-x}{x-3} \)
\(=\lim _{x \rightarrow 3^{+}} \frac{-(x-3)}{(x-3)}=-1 \)
\(\text { by }(1) \text { and }(2), f^{\prime}\left(3^{-}\right) \neq f^{\prime}\left(3^{+}\right)\)
\(\text { It is not differentiable }\)
\(\therefore f^{\prime}(x) \text { does not exist }\)

Given f is differentiable
\(\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 \)
\(f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}}-a(x+1) \)
\(=a(1+1)=2 a \)
\(\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} \)
\(=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1\)
\(\therefore 2 a =-1 \)
\(a =\frac{-1}{2} \)
\(\text { and } f(1)=1\)
\(a-b=1 \)
\(-1 / 2-1=b \)
\(b=-3 / 2 \)

\(f(x)=|x-1|+|x-3|+\sin x\)
\(\text { Since } \sin x \text { is differentiable everywhere }\)

At x = 1 and x = 3. The graph admit cups.
\(\therefore\) The derivative is not exist.
\(\therefore\) The number of points in R is 2.