Bonsoir,
Rappel de l'écriture de l'équation de la tangente:
[tex]\boxed{y = f'(a)(x-a)+f(a)}[/tex]
[tex]f(x) = \frac{8-x}{x^2+2}\\\\\\ f(0) = \frac{8-0}{0^2+2}\\\\
f(0) = \frac{8}{2}\\\\ \boxed{f(0) = 4}[/tex]
[tex]f(x) = \frac{8-x}{x^2+2} \, \text{ forme } (\frac{u}{v})'}\quad qui \, donne\; \boxed{\frac{u'v-v'u}{v^2}}\\\\
avec:\\
u = 8 - x \quad v = x^2+2\\
u' = -1 \quad v' = 2x\\\\
f'(x) = \frac{(-1)(x^2+2)-(2x)(8-x)}{(x^2+2)^2}\\\\
f'(x) = \frac{-x^2-2-16x+2x^2}{(x^2+2)^2}\\\\
\boxed{f'(x) = \frac {x^2-16x-2}{(x^2+2)^2}}\\\\\\ \text{ Maintenant que l'on a la d\'eriv\'ee en x on va pouvoir la faire en 0}\\\\[/tex]
[tex]f'(0) = \frac{0^2-16\times0-2}{(0^2+2)^2}\\\\
f'(0) = \frac{-2}{4}\\\\
\boxed{f'(0) = - \frac{1}{2}}[/tex]
[tex]y = f'(0)(x-0)+f(0)\\\\
y = - \frac{1}{2}\times x - (-\frac{1}{2}) \times 0 + 4\\\\
\boxed{y = -\frac{1}{2}x + 4}[/tex]
Bonne soirée :)
