Fusco Marcellini Sbordone Analisi Matematica 2 Esercizi Pdf 77 Upd May 2026

In some editions of the "Lite" version, Exercise 77 may refer to a Taylor expansion problem (e.g., "Write the Taylor series of the second order for a function...").

If your exercise asks for a Taylor expansion (Sviluppo di Taylor), here is the general method:

1. Formula: $$ f(x_0, y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) + \frac12[ \dots ] $$

2. Steps:

“Given the 1-form ( \omega = (e^x \sin y + yz)dx + (e^x \cos y + xz)dy + (xy + z)dz ), find the work done along a curve from (0,0,1) to (1,π,2).”
This tests exactness, potential functions, and path independence. In some editions of the "Lite" version, Exercise

While the exact text of Exercise 77 varies between editions (Zanichelli 2009, Liguori 2018, etc.), it commonly falls into one of these high-difficulty categories:

Liguori Editore (now part of Editoriale Scientifica) holds the copyright. No legal free PDF of the entire solution manual exists. However:

1. Define the Function and the Point of Interest Let the function be: $$ F(x, y) = y - x \sin(y) $$ We are analyzing the equation $F(x, y) = 0$ near the point $(x_0, y_0) = (0, 0)$. First, we verify that the point satisfies the equation: $$ F(0, 0) = 0 - 0 \cdot \sin(0) = 0 $$ The point lies on the level set.

2. Check the Regularity of $F$ We need to verify that $F$ is a $C^1$ function (continuously differentiable) in a neighborhood of $(0,0)$. We calculate the gradient $\nabla F(x, y)$: $$ \frac\partial F\partial x = -\sin(y) $$ $$ \frac\partial F\partial y = 1 - x \cos(y) $$ “Given the 1-form ( \omega = (e^x \sin

Both partial derivatives are combinations of sine, cosine, and polynomials, which are continuous functions everywhere in $\mathbbR^2$. Therefore, $F$ is of class $C^1$ (and actually $C^\infty$) in a neighborhood of the origin.

3. Apply the Implicit Function Theorem (Dini's Theorem) We want to solve for $y$ as a function of $x$ (i.e., $y = y(x)$). The theorem requires that the partial derivative with respect to the dependent variable ($y$) is non-zero at the point $(x_0, y_0)$.

Let's evaluate $\frac\partial F\partial y$ at $(0,0)$: $$ \left. \frac\partial F\partial y \right|_(0,0) = 1 - 0 \cdot \cos(0) = 1 - 0 = 1 $$

Since $\frac\partial F\partial y(0,0) = 1 \neq 0$, the hypotheses of the Implicit Function Theorem are satisfied. 1) to (1

4. Conclusion of the Theorem There exists a neighborhood $U$ of $x_0 = 0$ and a unique function $y: U \to \mathbbR$ such that:

5. Calculation of the Derivative We calculate the derivative of the implicit function $y'(x)$ using the formula provided by the theorem: $$ y'(x) = - \frac\frac\partial F\partial x\frac\partial F\partial y $$

We evaluate this at the point $x=0$ (knowing $y(0)=0$): $$ y'(0) = - \frac-\sin(y(0))1 - 0 \cdot \cos(y(0)) $$ $$ y'(0) = - \frac-\sin(0)1 - 0 $$ $$ y'(0) = - \frac01 = 0 $$

Result: The implicit function $y(x)$ exists locally. Its graph passes through the origin with a horizontal tangent (derivative is 0).