Questions in differentiation

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	A curve is given by the equations $x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,$ $y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta $, then the points for which $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,$ is given by
	If $y={{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n}},$ then $(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}$ is
	$f(x)$ and $g(x)$ are two differentiable function on $[0,\,2]$ such that $f''(x)-g''(x)=0,f'(1)=2,g'(1)=4$, $f(2)=3$, $g(2)=9,$ then $f(x)-g(x)$ at $x=3/2$ is
	If $y=a{{e}^{x}}+b{{e}^{-x}}+c$ where $a,b,c$ are parameters then ${{y}''}'=$
	If $y=a\cos \,(\log x)+b\sin \,(\log x)$ where $a,\,b$ are parameters then ${{x}^{2}}{y}''\,+\,x{y}'\,=$
	If $u={{x}^{2}}+{{y}^{2}}$ and $x=s+3t,y=2s-t,$ then $\frac{{{d}^{2}}u}{d{{s}^{2}}}=$
	$\frac{{{d}^{n}}}{d{{x}^{n}}}(\log x)$=
	The nth derivative of $x{{e}^{x}}$ vanishes when
	$\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\,\cos 3x)=$
	If $y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}-$....., then $\frac{{{d}^{2}}y}{d{{x}^{2}}}=$