Worked examples · Sheet 3

Second derivatives and local study of the implicit function

Theory used: Second derivative of the implicit function Local study of the implicit function Dini's theorem (two variables)

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Verify that the equation

x^{2} + y^{3} + y = 3

implicitly defines a function $y = φ (x)$ near the point $(1, 1)$ and compute $φ^{'} (1)$ and $φ^{''} (1)$ .

F (x, y) = x^{2} + y^{3} + y - 3, F (1, 1) = 1 + 1 + 1 - 3 = 0

F_{y} (1, 1) = 3 y^{2} + 1_{(1, 1)} = 4 \neq = 0

2 x + 3 y^{2} y^{'} + y^{'} = 0 ⟹ φ^{'} (1) = - \frac{2}{4} = - \frac{1}{2}

2 + 6 y (y^{'})^{2} + (3 y^{2} + 1) y^{''} = 0

2 + 6 \cdot \frac{1}{4} + 4 y^{''} = 0 ⟹ φ^{''} (1) = - \frac{7}{8}

The point belongs to the curve and $F_{y} (1, 1) = 4$ is nonzero, therefore Dini’s theorem applies; since $F$ is a polynomial, hence of class $C^{2}$ , the implicit function is of class $C^{2}$ . For the derivatives we differentiate the identity $x^{2} + φ (x)^{3} + φ (x) = 3$ with respect to $x$ : we obtain $2 x + (3 y^{2} + 1) y^{'} = 0$ , which at the point gives $2 + 4 φ^{'} (1) = 0$ , therefore $φ^{'} (1) = - \frac{1}{2}$ . Differentiating a second time, the term $3 y^{2} y^{'}$ produces $6 y (y^{'})^{2} + 3 y^{2} y^{''}$ by the product rule, hence the identity becomes $2 + 6 y (y^{'})^{2} + (3 y^{2} + 1) y^{''} = 0$ . Substituting $y = 1$ and $y^{'} = - \frac{1}{2}$ we obtain $2 + \frac{3}{2} + 4 y^{''} = 0$ , from which $φ^{''} (1) = - \frac{7}{8}$ .

Exercise 2

Verify that the equation

y^{3} + y + x^{2} - 2 x - 1 = 0

implicitly defines a function $y = φ (x)$ near the point $(1, 1)$ , and determine whether $x_{0} = 1$ is a local maximum or minimum point for $φ$ .

F (x, y) = y^{3} + y + x^{2} - 2 x - 1, F (1, 1) = 1 + 1 + 1 - 2 - 1 = 0

F_{y} (1, 1) = 3 y^{2} + 1_{(1, 1)} = 4 \neq = 0, F_{x} (1, 1) = 2 x - 2_{(1, 1)} = 0

φ^{'} (1) = - \frac{0}{4} = 0

(3 y^{2} + 1) y^{''} + 6 y (y^{'})^{2} + 2 = 0 ⟹ 4 φ^{''} (1) + 0 + 2 = 0

φ^{''} (1) = - \frac{1}{2} < 0 ⟹ x_{0} = 1 local maximum point

y = \varphi(x)

The curve

y^{3} + y + x^{2} - 2 x - 1 = 0

is the graph of

φ

: at

x = 1

the tangent is horizontal and

φ

has a local maximum

The hypotheses of Dini’s theorem are satisfied at the point, therefore the implicit function $φ$ exists and is of class $C^{2}$ , because $F$ is a polynomial. Since $F_{x} (1, 1) = 0$ , the first derivative $φ^{'} (1) = - \frac{F _{x}}{F _{y}}$ is zero, therefore $x_{0} = 1$ is a stationary point of $φ$ . For the second derivative we differentiate the identity $φ (x)^{3} + φ (x) + x^{2} - 2 x - 1 = 0$ twice: the first differentiation gives $(3 y^{2} + 1) y^{'} + 2 x - 2 = 0$ and the second gives $6 y (y^{'})^{2} + (3 y^{2} + 1) y^{''} + 2 = 0$ . At the point $y^{'} = 0$ , hence the first term is zero and what remains is $4 y^{''} + 2 = 0$ , that is $φ^{''} (1) = - \frac{1}{2}$ . Since the second derivative is negative at a stationary point, $x_{0} = 1$ is a local maximum point for the implicit function.

Exercise 3

Verify that the equation

e^{y} + y - x^{2} - 1 = 0

implicitly defines a function $y = φ (x)$ near the point $(0, 0)$ , study its monotonicity and determine the nature of the point $x_{0} = 0$ .

F (x, y) = e^{y} + y - x^{2} - 1, F (0, 0) = 1 + 0 - 0 - 1 = 0

F_{y} (x, y) = e^{y} + 1 > 0 for every (x, y)

φ^{'} (x) = - \frac{- 2 x}{e ^{y} + 1} = \frac{2 x}{e ^{φ (x)} + 1}

φ^{'} (x) < 0 for x < 0, φ^{'} (x) > 0 for x > 0

e^{y} (y^{'})^{2} + e^{y} y^{''} + y^{''} - 2 = 0 ⟹ 2 φ^{''} (0) - 2 = 0

φ^{''} (0) = 1 > 0 ⟹ x_{0} = 0 minimum point

The point $(0, 0)$ belongs to the curve and the derivative $F_{y} = e^{y} + 1$ is positive at every point of the plane, therefore Dini’s theorem applies and the implicit function is defined and of class $C^{2}$ near $0$ . The formula for the first derivative gives $φ^{'} (x) = \frac{2 x}{e ^{φ (x)} + 1}$ : the denominator is always positive, therefore the sign of $φ^{'}$ coincides with the sign of $x$ . Hence $φ$ is decreasing for $x < 0$ and increasing for $x > 0$ , and from this alone $x_{0} = 0$ turns out to be a minimum point. The second derivative confirms the result: differentiating the identity $e^{φ} + φ - x^{2} - 1 = 0$ twice we obtain $e^{y} (y^{'})^{2} + (e^{y} + 1) y^{''} - 2 = 0$ , and at the point, where $y = 0$ and $y^{'} = 0$ , what remains is $2 y^{''} - 2 = 0$ , that is $φ^{''} (0) = 1 > 0$ .