$$\eqalign{ & f'\left( {1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left[ {1 + h} \right]}}{{\left| {1 + h} \right|}} - \frac{{\left[ 1 \right]}}{{\left| 1 \right|}}}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{1 + h}} - 1}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h\left( {1 + h} \right)}} \cr & = - 1 \cr & f'\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left[ {1 - h} \right]}}{{\left| {1 - h} \right|}} - \frac{{\left[ 1 \right]}}{{\left| 1 \right|}}}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{0 - 1}}{{ - h}} \cr & = \infty \cr} $$

$$\eqalign{ & {x^2} + {t^2} = y \cr & \Rightarrow \frac{{dy}}{{dx}} = 2x + 2t.\frac{{dt}}{{dx}}......\left( 1 \right) \cr & {\text{As }}t = \frac{x}{{1 + {x^2}}} \Rightarrow \frac{{dt}}{{dx}} = \frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{Substitute these value of }}t{\text{ and }}\frac{{dt}}{{dx}}{\text{ in }}\left( 1 \right){\text{, we get}} \cr & \frac{{dy}}{{dx}} = 2x + \frac{{2x}}{{1 + {x^2}}}.\frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{On putting }}x = 2{\text{ in }}\frac{{dy}}{{dx}},{\text{ we get}} \cr & \frac{{dy}}{{dx}} = \frac{{488}}{{125}} \cr} $$

$$f\left( x \right) = {e^{ - x}}$$   satisfies $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$
It also satisfies $$f'\left( x \right) < 0$$   for all $$x.$$

Clearly, from the graph, $$f\left( x \right)$$  is non-differentiable at five points.

$$\eqalign{ & y = f\left( {{x^2}} \right)\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = f'\left( {{x^2}} \right).2x = 2x.\sqrt {2{{\left( {{x^2}} \right)}^2} - 1} \cr & {\text{At }}x = 1,\,\frac{{dy}}{{dx}} = 2.1.\sqrt {2 - 1} \cr} $$

$${\text{Let }}L = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2h + 2 + {h^2}} \right) - f\left( 2 \right)}}{{f\left( {h - {h^2} + 1} \right) - f\left( 1 \right)}}\,\,\,\,\,\,\,\left[ {\frac{0}{0}{\text{form}}} \right]$$
$$\therefore $$ Applying L'Hospital rule, we get
$$\eqalign{ & L = \mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {2h + 2 + {h^2}} \right).\left( {2 + 2h} \right)}}{{f'\left( {h - {h^2} + 1} \right).\left( {1 - 2h} \right)}} \cr & = \frac{{f'\left( 2 \right).2}}{{f'\left( 1 \right).1}} \cr & = \frac{{6 \times 2}}{{4 \times 1}} \cr & = 3 \cr} $$

$$\eqalign{ & f\left( x \right) = \min \left\{ {x + 1,\,\left| x \right| + 1} \right\} \cr & \Rightarrow f\left( x \right) = x + 1\,\forall \,x\, \in \,R \cr} $$

Hence, $$f\left( x \right)$$  is differentiable everywhere for all $$x\, \in \,R.$$

Let $$y = \log \,x$$
$$ \Rightarrow {y_1} = \frac{1}{x},\,{y_2} = \frac{{ - 1}}{{{x^2}}},\,{y_3} = \frac{2}{{{x^3}}},.....,\,{y_n} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}}$$

By mean value theorem, there exists a real number $$c\, \in \left( {2,\,4} \right)$$   such that
$$\eqalign{ & f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 2 \right)}}{{4 - 2}} \cr & \Rightarrow f'\left( c \right) = \frac{{f\left( 4 \right) + 4}}{2} \cr & {\text{Since, }}f'\left( c \right) \geqslant 6,\,\forall \,x\, \in \left[ {2,\,4} \right] \cr & \therefore \,f'\left( c \right) \geqslant 6 \cr & \Rightarrow \frac{{f\left( 4 \right) + 4}}{2} \geqslant 6 \cr & \Rightarrow f\left( 4 \right) + 4 \geqslant 12 \cr & \Rightarrow f\left( 4 \right) \geqslant 8 \cr} $$

$$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\,;$$
As function is differentiable so it is continuous as it is given that $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$     and hence $$f\left( 1 \right) = 0$$
Hence, $$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right)}}{h} = 5$$