Fusco Marcellini Sbordone Analisi Matematica 2 Esercizi Pdf 77 May 2026

By following these steps, you should be able to find or create a useful study resource for "Analisi Matematica 2" by Fusco, Marcellini, and Sbordone.

This is a fictional, in-depth academic commentary inspired by the type of exercise you might find on page 77 of a theoretical exercise collection such as Fusco, Marcellini, Sbordone – Analisi Matematica 2 – Esercizi e problemi. Since the actual PDF is copyrighted and not reproducible here, the following is a reconstructed analytical problem (in the style of that textbook) along with its detailed step-by-step solution, focusing on multivariable calculus and ordinary differential equations — typical topics in Analisi 2.

The authors are known for their rigor and for bridging theory with subtle counterexamples. Page 77 likely falls within the chapters on limits and continuity in multiple variables, or differential calculus for vector functions, possibly the initial part of implicit functions or double integrals.

Below is a representative exercise — thematically plausible for that page — followed by a full analytical solution.

Let ( f: \mathbbR^2 \to \mathbbR ) be defined as:

[ f(x,y) = \begincases \dfracx^3 + y^3x^2 + y^2, & (x,y) \neq (0,0) \ 0, & (x,y) = (0,0) \endcases ]

If "Esercizi Pdf 77" refers to a specific set of exercises:

Let ( v = (\cos\theta, \sin\theta) ). Then: By following these steps, you should be able

[ D_v f(0,0) = \lim_t \to 0 \fracf(t\cos\theta, t\sin\theta) - f(0,0)t = \lim_t \to 0 \fract^3(\cos^3\theta + \sin^3\theta)/t^2t = \lim_t \to 0 \fract(\cos^3\theta + \sin^3\theta)t = \cos^3\theta + \sin^3\theta. ]

Notice: For ( \theta=0 ), we get ( 1 ) (matches ( f_x )), for ( \theta=\pi/2 ) we get ( 1 ) (matches ( f_y )), but generally ( D_v f(0,0) ) equals ( v \cdot \nabla f(0,0) ) only if ( \cos^3\theta+\sin^3\theta = \cos\theta+\sin\theta ), which is false for most ( \theta ) (e.g., ( \theta=45^\circ ): LHS ( \sqrt2/2 ), RHS ( \sqrt2 )).
This confirms non-differentiability (directional derivative is not linear in ( v )).

We compute ( f_x(x,y) ) for ( (x,y) \neq (0,0) ):

[ f_x(x,y) = \frac\partial\partial x \left( \fracx^3 + y^3x^2 + y^2 \right) = \frac3x^2(x^2+y^2) - (x^3+y^3)(2x)(x^2+y^2)^2. ]

Simplify: ( \frac3x^4 + 3x^2y^2 - 2x^4 - 2xy^3(x^2+y^2)^2 = \fracx^4 + 3x^2y^2 - 2xy^3(x^2+y^2)^2 ).

Along the line ( y = x ):

[ f_x(x,x) = \fracx^4 + 3x^4 - 2x^4(2x^2)^2 = \frac2x^44x^4 = \frac12. ] The authors are known for their rigor and

But ( f_x(0,0) = 1 ). So ( f_x ) is not continuous at ( (0,0) ). Similarly for ( f_y ).

The candidate derivative is the linear map ( L(h,k) = h + k ) (since gradient ( \nabla f(0,0) = (1,1) )).

We must check:

[ \lim_(h,k) \to (0,0) \fracf(h,k) - f(0,0) - (h + k)\sqrth^2 + k^2 = 0. ]

Compute: ( f(h,k) = \frach^3 + k^3h^2 + k^2 ). Then:

[ f(h,k) - (h + k) = \frach^3 + k^3h^2 + k^2 - (h + k) = \frach^3 + k^3 - (h+k)(h^2 + k^2)h^2 + k^2. ]

Expanding: ( (h+k)(h^2 + k^2) = h^3 + hk^2 + kh^2 + k^3 ). Thus: Let ( f: \mathbbR^2 \to \mathbbR ) be

Numerator ( h^3 + k^3 - h^3 - hk^2 - kh^2 - k^3 = -hk(h + k) ).

Therefore:

[ \fracf(h,k) - (h + k)\sqrth^2 + k^2 = \frac-hk(h + k)(h^2 + k^2)^3/2. ]

Now use polar ( h = r\cos\theta, k = r\sin\theta ):

[ \frac-r^2\cos\theta\sin\theta \cdot r(\cos\theta+\sin\theta)r^3 = -\cos\theta\sin\theta(\cos\theta+\sin\theta). ]

This expression depends on ( \theta ), not just ( r ). For ( \theta = \pi/4 ), we get ( -(\frac12)(\sqrt2) \neq 0 ).
Hence the limit does not exist (fails to be 0). So ( f ) is not differentiable at ( (0,0) ), despite having partial derivatives.

The inclusion of "Pdf" in the query highlights the modern student's reliance on digital formats. The physical textbook is a staple in Italian university bookstores (often published by Zanichelli or Editori Riuniti), but the PDF format is sought after for several reasons: