1.

$f'_1(x)=(x\sin x)'=x'\sin x+x(\sin x)'=\sin x+x\cos x$

$f'_2(x)=\dfrac{(\cos x)' x-x'\cos x}{x^2}=\dfrac{-x\sin x -\cos x}{x^2}$

$\to \dfrac{f'_2(x)}{f'_1(x)}=\dfrac{-x\sin x-\cos x}{x^2(\sin x+x\cos x}$

$\to \dfrac{f'_2(1)}{f'_1(1)}=-1$

2.

$y=\dfrac{\tan\sqrt{x}}{\sqrt{x}}$

$dy=\dfrac{(\tan\sqrt{x})'\sqrt{x}- \tan\sqrt{x}.(\sqrt{x})'}{x} dx$

$=\dfrac{\dfrac{1}{2\cos^2\sqrt{x}}-\tan\sqrt{x}.\dfrac{1}{2\sqrt{x}}}{x}dx$

$=\dfrac{1}{2x\cos^2\sqrt{x}}-\dfrac{\tan\sqrt{x}}{2x\sqrt{x}}dx$

3.

$y'=3(x^2+1)^2(x^2+1)'$

$=3(x^2+1)^2.2x$

$=6x(x^2+1)^2$

$y''=6(x^2+1)^2+6x[(x^2+1)^2]' $

$=6(x^2+1)^2+2.6x(x^2+1)(x^2+1)'$

$=6(x^2+1)^2+24x^2(x^2+1)$

$=6(x^4+2x^2+1)+24x^4+24x^2$

$=30x^4+36^2+6$

$y'''=120^3+72x$

Giải thích các bước giải:

 Ta có:

1,

\(\begin{array}{l}
{f_1}\left( x \right) = x.\sin x\\
 \Rightarrow {f_1}'\left( x \right) = x'.\sin x + x.\left( {\sin x} \right)' = \sin x + x\cos x\\
 \Rightarrow {f_1}'\left( 1 \right) = \sin 1 + 1.\cos 1 = \sin 1 + \cos 1\\
{f_2}\left( x \right) = \dfrac{{\cos x}}{x}\\
 \Rightarrow {f_2}'\left( x \right) = \dfrac{{\left( {\cos x} \right)'.x - x'.\cos x}}{{{x^2}}} = \dfrac{{ - \sin x.x - \cos x}}{{{x^2}}}\\
 \Rightarrow {f_2}'\left( 1 \right) = \dfrac{{ - \sin 1.1 - \cos 1}}{{{1^2}}} =  - \left( {\sin 1 + \cos 1} \right)\\
 \Rightarrow \dfrac{{{f_1}'\left( 1 \right)}}{{{f_2}'\left( 1 \right)}} =  - 1
\end{array}\)

2,

\(\begin{array}{l}
y = \dfrac{{\tan \sqrt x }}{{\sqrt x }}\\
 \Rightarrow y' = \dfrac{{\left( {\tan \sqrt x } \right)'.\sqrt x  - \sqrt x '.tan\sqrt x }}{x}\\
 = \dfrac{{\dfrac{{\left( {\sqrt x } \right)'}}{{{{\cos }^2}\sqrt x }}.\sqrt x  - \dfrac{1}{{2\sqrt x }}.\tan \sqrt x }}{x}\\
 = \dfrac{{\dfrac{1}{{2\sqrt x .{{\cos }^2}\sqrt x }}.\sqrt x  - \dfrac{{\tan \sqrt x }}{{2\sqrt x }}}}{x}\\
 = \dfrac{{\dfrac{1}{{2{{\cos }^2}\sqrt x }} - \dfrac{{\tan \sqrt x }}{{2\sqrt x }}}}{x}\\
 = \dfrac{{\dfrac{{\sqrt x  - \tan \sqrt x .co{s^2}\sqrt x }}{{2\sqrt x {{\cos }^2}\sqrt x }}}}{x}\\
 = \dfrac{{\sqrt x  - \sin \sqrt x .\cos \sqrt x }}{{2x\sqrt x {{\cos }^2}\sqrt x }}\\
 \Rightarrow dy = \dfrac{{\sqrt x  - \sin \sqrt x .\cos \sqrt x }}{{2x\sqrt x {{\cos }^2}\sqrt x }}dx\\
3,\\
y = {\left( {{x^2} + 1} \right)^3} = {x^6} + 3{x^4} + 3{x^2} + 1\\
 \Rightarrow y' = 6{x^5} + 3.4{x^3} + 3.2x = 6{x^5} + 12{x^3} + 6x\\
 \Rightarrow y'' = 6.5{x^4} + 12.3{x^2} + 6 = 30{x^4} + 36{x^2}\\
 \Rightarrow y''' = 30.4{x^3} + 36.2x = 120{x^3} + 72x
\end{array}\)