$$\because \,y = {\log ^n}x$$
On differentiating w.r.t. $$x,$$ we get
$$\eqalign{ & x\,\log \,x\,{\log ^2}x\,{\log ^3}x.....{\log ^{n - 1}}x\,{\log ^n}x\frac{{dy}}{{dx}} \cr & = \frac{{x\,\log \,x\,{{\log }^2}x\,{{\log }^3}x.....{{\log }^{n - 1}}x\,{{\log }^n}x.1}}{{x\,\log \,x\,{{\log }^2}x\,{{\log }^3}x.....{{\log }^{n - 1}}x}} \cr & = {\log ^n}x \cr} $$

Given that $$f:R \to R$$   such that
$$\eqalign{ & f\left( 1 \right) = 3\,\,{\text{and}}\,\,f'\left( 1 \right) = 6 \cr & {\text{Then}}\,\,\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{f\left( {1 + x} \right)}}{{f\left( 1 \right)}}} \right)^{\frac{1}{x}}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \,\frac{1}{x}\left[ {\log \,f\left( {1 + x} \right) - \log \,f\left( 1 \right)} \right]}} \cr & = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{f\left( {1 + x} \right)}}f'\left( {1 + x} \right)}}{1}}}\,\,\,\,\left[ {{\text{Using L'Hospital rule}}} \right] \cr & = {e^{\frac{{f'\left( 1 \right)}}{{f\left( 1 \right)}}}} \cr & = {e^{\frac{6}{3}}} \cr & = {e^2} \cr} $$

$$\eqalign{ & \frac{{dx}}{{dy}} = \frac{{\frac{{dx}}{{dt}}}}{{\frac{{dy}}{{dt}}}} = \frac{{\cos \,t - t\sin \,t}}{{1 + \cos \,t}} \cr & \therefore \frac{{{d^2}x}}{{d{y^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dx}}{{dy}}} \right)}}{{\frac{{dy}}{{dt}}}} \cr & = \frac{{\frac{{\left( { - 2\sin \,t - t\cos \,t} \right)\left( {1 + \cos \,t} \right) - \left( {\cos \,t - t\sin \,t} \right)\left( { - \sin \,t} \right)}}{{{{\left( {1 + \cos \,t} \right)}^2}}}}}{{1 + \cos \,t}} \cr & {\text{Now, put }}t = \frac{\pi }{2} \cr} $$

$$\eqalign{ & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & \left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}} \right| \leqslant \mathop {\lim }\limits_{h \to 0} \left| {\frac{{{{\left( h \right)}^2}}}{h}} \right| \cr & \Rightarrow \left| {f'\left( x \right)} \right| \leqslant 0 \cr & \Rightarrow f'\left( x \right) = 0 \cr & \Rightarrow f\left( x \right) = {\text{constant}} \cr & {\text{As }}f\left( 0 \right) = 0\,\, \Rightarrow f\left( 1 \right) = 0 \cr} $$

$$\eqalign{ & {\text{Let }}g\left( x \right) = {f^3}\left( x \right) \cr & \Rightarrow g'\left( x \right) = 3{f^2}\left( x \right).f'\left( x \right) \cr & \because \,f:\left[ {2,\,7} \right] \to \left[ {0,\,\infty } \right) \Rightarrow g:\left[ {2,\,7} \right] \to \left[ {0,\,\infty } \right) \cr} $$
Using Lagrange's mean value theorem on $$g\left( x \right),$$   we get $$g'\left( c \right) = \frac{{g\left( 7 \right) - g\left( 2 \right)}}{5},\,c\, \in \left[ {2,\,7} \right]$$
$$ \Rightarrow 2{f^2}\left( c \right)f'\left( c \right) = \left( {f\left( 7 \right) - f\left( 2 \right)} \right)\frac{{{{\left( {f\left( 7 \right)} \right)}^2} + {{\left( {f\left( 2 \right)} \right)}^2} + f\left( 2 \right)f\left( 7 \right)}}{3}$$

$$\eqalign{ & f\left( x \right) = \root 3 \of {\frac{{{x^4}}}{{\left| x \right|}}} ,\,x \ne 0,\,f\left( 0 \right) = 0 \cr & \therefore \,f\left( x \right) = \root 3 \of {\frac{{{x^4}}}{{ - x}}} = \root 3 \of {{ - x^3}} = - x{\text{ if }}x < 0 \cr & \& \,f\left( x \right) = \root 3 \of {\frac{{{x^4}}}{x}} = \root 3 \of {{x^3}} = x{\text{ if }}x > 0 \cr} $$
\[f\left( x \right) = \left\{ \begin{array}{l} - x,{\rm{ \,if\, }}x < 0\\ \,\,\,\,\,0,{\rm{ \,if\, }}x = 0\\ \,\,\,\,\,x,{\rm{ \,if\, }}x > 0\,\, \end{array} \right.\]
Clearly $$f\left( x \right)$$  is continuous for all $$x$$ but not differentiable at $$x = 0$$

$$\eqalign{ & {\text{Since, }}f\left( x \right) = 15 - \left| {\left( {10 - x} \right)} \right| \cr & \therefore g\left( x \right) = f\left( {f\left( x \right)} \right) = 15 - \left| {10 - \left[ {15 - \left| {10 - x} \right|} \right]} \right| \cr & = 15 - \left| {\left| {10 - x} \right| - 5} \right| \cr} $$
$$\therefore $$ Then, the points where function $$g\left( x \right)$$  is
Non-differentiable are
$$\eqalign{ & 10 - x = 0\,\,{\text{and }}\left| {10 - x} \right| = 5 \cr & \Rightarrow x = 10\,\,{\text{and }}x - 10 = \pm 5 \cr & \Rightarrow x = 10\,\,{\text{and }}x = 15,\,5 \cr} $$

$$\eqalign{ & y = \frac{{{{\left( {a - x} \right)}^{\frac{3}{2}}} + {{\left( {x - b} \right)}^{\frac{3}{2}}}}}{{\sqrt {a - x} + \sqrt {x - b} }} \cr & = \frac{{\left( {\sqrt {a - x} + \sqrt {x - b} } \right)\left( {a - x - \sqrt {a - x} \sqrt {x - b} + x - b} \right)}}{{\sqrt {a - x} + \sqrt {x - b} }} \cr & = a - b - \sqrt {a - x} \sqrt {x - b} \cr & {\text{or }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {a - x} }}\sqrt {x - b} - \frac{1}{{2\sqrt {x - b} }}\sqrt {a - x} \cr & = \frac{{2x - a - b}}{{2\sqrt {a - x} \sqrt {x - b} }} \cr} $$

Given that $$f\left( x \right)$$  is differentiable on $$\left( {0,\,\infty } \right)$$  with
$$f\left( 1 \right) = 1\,{\text{and}}\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f\left( x \right) - {x^2}f\left( t \right)}}{{t - x}} = 1$$         for each $$x>0$$
$$\eqalign{ & \Rightarrow \mathop {\lim }\limits_{t \to x} \frac{{2t\,f\left( x \right) - {x^2}f'\left( t \right)}}{1} = 1\,\,\left[ {{\text{using L'Hospital rule}}} \right] \cr & \Rightarrow 2x\,f\left( x \right) - {x^2}\,f'\left( x \right) = 1 \cr & \Rightarrow f'\left( x \right) - \frac{2}{x}f\left( x \right) = - \frac{1}{{{x^2}}}\left[ {{\text{Linear differential equation}}} \right] \cr} $$
Integrating factor
$$\eqalign{ & {e^{\int { - \,\frac{2}{x}dx} }} = {e^{ - 2\,\log \,x}} = {e^{\log \frac{1}{{{x^2}}}}} = \frac{1}{{{x^2}}} \cr & \therefore \,\,{\text{Solution is}}\,f\left( x \right) \times \frac{1}{{{x^2}}} = \int {\left( { - \frac{1}{{{x^2}}}} \right) \times \frac{1}{{{x^2}}}dx} \cr & \Rightarrow \frac{{f\left( x \right)}}{{{x^2}}} = \frac{1}{{3{x^3}}} + C \cr & \Rightarrow f\left( x \right) = C{x^2} + \frac{1}{{3x}} \cr & {\text{Also }}f\left( 1 \right) = 1 \cr & \Rightarrow 1 = C + \frac{1}{3} \cr & \Rightarrow C = \frac{2}{3} \cr & \therefore f\left( x \right) = \frac{2}{3}{x^2} + \frac{1}{{3x}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{g\left( {a + h} \right)f\left( a \right) - g\left( a \right)f\left( {a + h} \right)}}{h}\left[ {{\text{For }}x = a + h} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{g\left( {a + h} \right)f\left( a \right) - g\left( a \right)f\left( a \right) + g\left( a \right)f\left( a \right) - g\left( a \right)f\left( {a + h} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} f\left( a \right)\left[ {\frac{{g\left( {a + h} \right) - g\left( a \right)}}{h}} \right] - \mathop {\lim }\limits_{h \to 0} g\left( a \right)\left[ {\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}} \right] \cr & = f\left( a \right)g'\left( a \right) - g\left( a \right)f'\left( a \right) \cr & = 2 \times 2 - \left( { - 1} \right) \times 1 \cr & = 5 \cr} $$