$$\eqalign{ & y = {\log _{10}}x + {\log _x}10 + {\log _x}x + {\log _{10}}10 \cr & y = {\log _{10}}x + {\log _x}10 + 1 + 1 \cr} $$
Differentiating equation w.r.t. $$x$$
$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{x\,{{\log }_e}10}} - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}}}.\frac{1}{{\left( {x\,\log \,10} \right)}} \cr & = \frac{1}{{x\,{{\log }_e}10}}\left[ {1 - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}}}} \right] \cr & {\left( {\frac{{dy}}{{dx}}} \right)_{x = 10}} = \frac{1}{{10\,{{\log }_e}10}}\left[ {1 - 1} \right] = 0 \cr} $$

\[\left[ \begin{array}{l} {\bf{Note :}}\\ {\log _x}10 = \frac{{{{\log }_{10}}10}}{{{{\log }_{10}}x}} = \frac{1}{{{{\log }_{10}}x}}\\ \frac{d}{{dx}}\left[ {\frac{1}{{{{\log }_{10}}x}}} \right] = - {\left( {{{\log }_{10}}x} \right)^{ - 2}} \times \frac{1}{{x\,{{\log }_e}10}}\\ = - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}x\,{{\log }_e}10}} \end{array} \right]\]

Let $$y = {\left( {1 + \frac{1}{x}} \right)^x}$$
Taking logarithm of both sides, we get
$$\eqalign{ & \log \,y = x\left[ {\log \left( {1 + \frac{1}{x}} \right)} \right] \cr & \Rightarrow \frac{1}{y}{y_1}\left( x \right) = \frac{{{x^2}}}{{x + 1}}\left( { - \frac{1}{{{x^2}}}} \right) + \log \left( {1 + \frac{1}{x}} \right) \cr & \Rightarrow - \frac{1}{{x + 1}} + \log \left( {1 + \frac{1}{x}} \right).....(1) \cr & {\text{Since, }}y\left( 2 \right) = {\left( {1 + \frac{1}{2}} \right)^2} = \frac{9}{4} \cr & {\text{So, }}{y_1}\left( 2 \right) = \left( {\frac{9}{4}} \right)\left( { - \frac{1}{3} + \log \frac{3}{2}} \right) \cr} $$
Again differentiate equation (1) w.r.t. $$\left( x \right),$$ we get
$$\frac{{y\left( x \right){y_2}\left( x \right) - {{\left[ {{y_1}\left( x \right)} \right]}^2}}}{{{{\left( {y\left( x \right)} \right)}^2}}} = \frac{1}{{{{\left( {x + 1} \right)}^2}}} - \frac{1}{{x\left( {x + 1} \right)}}$$
By putting $$x = 2,$$  we get
$$\frac{{y\left( 2 \right){y_2}\left( 2 \right) - {{\left( {{y_1}\left( 2 \right)} \right)}^2}}}{{{{\left( {y\left( 2 \right)} \right)}^2}}} = \frac{ - 1}{{18}}$$
Now, put value of $${y\left( 2 \right)}$$  and $${{y_1}\left( 2 \right)}$$
$$\eqalign{ & \Rightarrow \,{y_2}\left( 2 \right) = \left( {\frac{9}{4}} \right){\left( { - \frac{1}{3} + \log \frac{3}{2}} \right)^2} - \frac{1}{8}{\left( {{y_2}\left( 2 \right) + \frac{1}{8}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9{\left( {\log \frac{3}{2} - \frac{1}{3}} \right)^2} \cr & \Rightarrow {\text{ Required expression }} = 3 \cr} $$

We have \[f\left( x \right) = \left\{ \begin{array}{l} \left( {x - 1} \right)\sin \frac{1}{{x - 1}}\,{\rm{if}}\,\,x \ne 1\\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x = 1 \end{array} \right.\]
$$\eqalign{ & Rf\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{h\,\sin \frac{1}{h} - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \,\sin \frac{1}{h} = {\text{a finite number}} \cr} $$
Let this finite number be $$l$$
$$\eqalign{ & Lf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - h\,\sin \left( {\frac{1}{{ - h}}} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \,\sin \left( {\frac{1}{{ - h}}} \right) \cr & = - \mathop {\lim }\limits_{h \to 0} \,\sin \left( {\frac{1}{h}} \right) = - \left( {{\text{a finite number}}} \right) = - l \cr} $$
Thus $$Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$$
$$\therefore f$$ is not differentiable at $$x = 1$$
Also,
$$\eqalign{ & {\left. {f'\left( 0 \right) = \sin \frac{1}{{\left( {x - 1} \right)}} - \frac{{x - 1}}{{{{\left( {x - 1} \right)}^2}}}\cos \left( {\frac{1}{{x - 1}}} \right)} \right]_{x = 0}} \cr & = - \sin \,1 + \cos \,1 \cr} $$
$$\therefore f$$ is differentiable at $$x =0$$

$$\eqalign{ & f\left( { - x} \right) = f\left( x \right) \Rightarrow - f'\left( { - x} \right) = f'\left( x \right) \cr & \therefore \,f'\left( x \right){\text{ is odd}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Let }}x = \tan \,\theta \cr & {\text{Then }}{\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x} = {\tan ^{ - 1}}\frac{{\sec \,\theta - 1}}{{\tan \,\theta }} \cr & = {\tan ^{ - 1}}\frac{{1 - \cos \,\theta }}{{\sin \,\theta }} \cr & = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x \cr & \therefore \frac{{d\left( {{{\tan }^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)}}{{d\left( {{{\tan }^{ - 1}}x} \right)}} = \frac{1}{2} \cr} $$

Given expression can be written as
$$\eqalign{ & y = {\tan ^{ - 1}}\left[ {\frac{{{2^x}\left( {2 - 1} \right)}}{{1 + {2^x}{{.2}^{2x + 1}}}}} \right] \cr & \,\,\,\,\, = {\tan ^{ - 1}}\left[ {\frac{{{2^{x + 1}} - {2^x}}}{{1 + {2^x}{{.2}^{2x + 1}}}}} \right] \cr & \,\,\,\,\, = {\tan ^{ - 1}}\left( {{2^{x + 1}}} \right) - {\tan ^{ - 1}}\left( {{2^x}} \right) \cr & \Rightarrow \,\frac{{dy}}{{dx}} = \frac{{{2^{x + 1}}\log \,2}}{{1 + {2^{2\left( {x + 1} \right)}}}} - \frac{{{2^x}\log \,2}}{{1 + {2^{2x}}}} \cr & \therefore \,{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = \left( {\log \,2} \right)\left( {\frac{2}{5} - \frac{1}{2}} \right) = \log \,2\left( { - \frac{1}{{10}}} \right) \cr} $$

Given that $$f\left( x \right)$$  is a continuous and differentiable function and $$f\left( {\frac{1}{x}} \right) = 0,\,x = n,\,n \in \,I$$
$$\therefore f\left( {{0^ + }} \right) = f\left( {\frac{1}{\infty }} \right) = 0$$
Since R.H.L. $$=0$$
$$\therefore f\left( 0 \right) = 0$$   for $$f\left( x \right)$$  to be continuous.
$$\eqalign{ & {\text{Also }}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - f\left( 0 \right)}}{{h - 0}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} = 0 \cr & = 0\left[ {{\text{Using }}f\left( 0 \right) = 0{\text{ and }}f\left( {{0^ + }} \right) = 0} \right] \cr & {\text{Hence }}f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0 \cr} $$

$$\eqalign{ & {\text{RH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\lambda + \mu \left| {0 + h} \right| + \nu {{\left| {0 + h} \right|}^2} - \lambda }}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\mu h + \nu {h^2}}}{h} \cr & = \mu \cr & {\text{LH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\lambda + \mu \left| {0 - h} \right| + \nu {{\left| {0 - h} \right|}^2} - \lambda }}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\mu h + \nu {h^2}}}{{ - h}} \cr & = - \mu \cr & f'\left( x \right)\,{\text{exists}} \Rightarrow {\text{RH derivative}} = {\text{LH derivative}} \Rightarrow \mu = - \mu \cr} $$

$$\frac{{du}}{{dv}} = \frac{{\frac{{du}}{{dx}}}}{{\frac{{dv}}{{dx}}}} = \frac{{f'\left( {{x^3}} \right).3{x^2}}}{{g'\left( {{x^2}} \right).2x}} = \frac{{\cos \,{x^3}.3{x^2}}}{{\sin \,{x^2}.2x}} = \frac{3}{2}x\,\cos \,{x^3}.{\text{cosec}}\,{x^2}$$

$$\eqalign{ & {\text{For }}x \ne 0,{\text{ we have}} \cr & f\left( x \right) = x + \frac{{\frac{x}{{1 + x}} + x}}{{1 - \frac{1}{{1 + x}}}} = x + \frac{{\frac{x}{{1 + x}}}}{{\frac{x}{{1 + x}}}} = x + 1 \cr & {\text{For }}x = 0,\,f\left( x \right) = 0 \cr} $$
Thus, \[f\left( x \right) = \left\{ \begin{array}{l} x + 1,\,\,\,\,\,x \ne 0\\ \,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \end{array} \right.\]
$${\text{Clearly, }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 \ne f\left( 0 \right)$$
So, $$f\left( x \right)$$  is discontinuous and hence not differentiable at $$x = 0.$$