Bonjour j'aurais besoin d'aide svp à cette question svp​

[tex]h(x) = \frac{2x}{x {}^{2} + 9} = \frac{u(x)}{v(x)} [/tex]

avec u(x) = 2x et v(x) = x² + 9

u'(x) = 2

v'(x) = 2x

h'(x) = ( u'(x)v(x) - u(x)v'(x) ) / v²(x)

h'(x) =

[tex] \frac{2(x {}^{2} + 9) - 2x \times 2x}{(x {}^{2} + 9) {}^{2} } [/tex]

=

[tex] \frac{2x {}^{2} + 18 - 4x {}^{2} }{(x {}^{2} + 9) {}^{2} } [/tex]

=

[tex] \frac{ - 2x {}^{2} + 18 }{(x {}^{2} + 9) {}^{2} } [/tex]

=

[tex] \frac{2( - x {}^{2} + 9)}{(x {}^{2} + 9) {}^{2} } [/tex]

=

[tex] \frac{2(3 + x)(3 - x)}{(x {}^{2} + 9) {}^{2} } [/tex]