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Esercizi svolti di analisi reale e complessa - Dipartimento di ...

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f(0) = 0 <br />

0 = F ′ (0) =<br />

<br />

2x + (cosh f(x))f ′ (x) + e xf(x) (f(x) + xf ′ (x))<br />

= f ′ (0)<br />

x = 0 <br />

<br />

0 = F ′′ <br />

(0) = 2 + (sinh f(x))(f ′ (x)) 2 + (cosh f(x))f ′′ (x) +<br />

+ e xf(x) (f(x) + xf ′ (x)) 2 + 2f ′ (x) + xf ′′ (x) <br />

=<br />

|x=0<br />

= 2 + f ′′ (0) ,<br />

f ′′ (0) = −2 <br />

<br />

<br />

<br />

<br />

∂f<br />

∂x =<br />

∂ F<br />

∂ y (0, 0) = [cosh y + xexy ] |(x,y)=(0,0) =<br />

⎛<br />

⎜<br />

⎝<br />

∂f1<br />

∂x1<br />

<br />

∂f6<br />

∂x1<br />

= 1 = 0.<br />

∂f1<br />

∂x2<br />

<br />

∂f6<br />

∂x2<br />

⎞<br />

⎟<br />

⎠ ∂f<br />

∂y =<br />

⎛<br />

∂<br />

⎜<br />

f(x, g(t)) = ⎜<br />

∂t ⎝<br />

<br />

⎛<br />

⎜<br />

⎝<br />

∂f1<br />

∂y1<br />

<br />

∂f6<br />

∂y1<br />

∂f1<br />

∂y2<br />

<br />

∂f6<br />

∂y2<br />

3 ∂f1<br />

j=1 ∂yj g′ j (t)<br />

<br />

3 ∂f6<br />

j=1 ∂yj g′ j (t)<br />

⎞<br />

⎟<br />

⎠<br />

∂f1<br />

∂y3<br />

<br />

∂f1<br />

∂y3<br />

⎞<br />

⎟<br />

⎠<br />

<br />

|x=0<br />

=

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