6. Implicit functions

If a function is given by the expression y = f (x), i.e. the value of the dependent variable y is given by the formula in the independent variable x, we say the function f is given explicitly ( f is an explicit function). For instance, the linear function y = 7x + 3 is an explicit function. If the dependence of y on x is given by an equation in both x and y, for instance x2 + y2 = 1, we say the function y(x) is given implicitly (y is an implicit function). However, not every equation in x and y defines an implicit function y(x), for instance x2 + y2 + 1 = 0.

6.1 Implicit functions of one variable Definition 6.1.1 An equation F(x,y) = 0 defines an implicit function y = f (x) on a neighbourhood of a point (x0,y0) if 1. F(x0,y0) = 0, 2. ∃δ > 0,ε > 0 such that ∀x ∈ Oδ (x0) ∃!y ∈ Oε (y0) : F(x,y) = 0.

R The function y = f (x) is implicitly defined by the equation F(x,y) = 0 in the neighbourhood of the point (x0,y0). Thus, • f is a function of one real variable, • f is defined for all x ∈ (x0 − δ,x0 + δ) for a δ > 0, • the values y = f (x) lie within the interval (y0 − ε,y0 + ε) for an ε > 0, • for an x ∈ (x0 − δ,x0 + δ) there exists exactly one value y = f (x) ∈ (y0 − ε,y0 + ε) such that the equality F(x,y) = 0 holds, i.e. the solution to the equation F(x,y) = 0 within the rectangle (x0 − δ,x0 + δ) × (y0 − ε,y0 + ε) is the set {(x, f (x)) : x ∈ (x0 − δ,x0 + δ)} and thus

F(x, f (x)) = 0 on (x0 − δ,x0 + δ),

• f (x0) = y0.

2 k Theorem 6.1.1 Let G ⊆ R be an open set and let F ∈ C (G) for k ∈ N. Let (x0,y0) ∈ G be a point such that ∂F F(x ,y ) = 0 and (x ,y ) 6= 0. 0 0 ∂y 0 0

Then the equation F(x,y) = 0 defines an implicit function y = f (x) on a neighbourhood of a point (x0,y0). k Moreover, f ∈ C (x0 − δ,x0 + δ) for a δ > 0. 66 Chapter 6. Implicit functions

R • The existence and continuity of higher order partial derivatives of F implies the existence and k continuity of higher order derivatives of f . More precisely, if F ∈C ((x0 −δ,x0 +δ)×(y0 −ε,y0 +ε)) k then f ∈ C (x0 − δ,x0 + δ). • The existence of a point (x0,y0) such that F(x0,y0) = 0 is necessary for the existence of an implicit function y = f (x). It specifies a point lying on the graph of f . If such (x0,y0) did not exist, there would not exist a function y = f (x) implicitly defined by F(x,y) = 0, i.e. satisfying F(x, f (x)) = 0. • The graph of implicitly defined function is a subset of the contour line of the graph of F(x,y) ∂F corresponding to the cutting plane z = 0. The condition (x ,y ) 6= 0 implies this contour line is the ∂y 0 0 graph of a function in x in a neighbourhood of the point (x0,y0). Hence, the equation F(x,y) = 0 has, except the solution (x0,y0), infinitely many solutions of the form (x, f (x)). The set of these solutions is the graph of the implicit function f .

Figure 6.1: Relation between the graphs of f and F

Theorem 6.1.1 is referred to as the theorem on the existence of implicit functions. It does not say anything about how large δ and ε in Definition 6.1.1 could be, the explicit expression of the implicit function is neither specified. Often it is even impossible to determine explicit expressions of implicit functions.

 Example 6.1 Let us consider the equation

F(x,y) = x2 + y2 − 1 = 0 ∂F describing a circle. Because F(0,1) = 0, the point (0,1) lies on the circle. Because (0,1) = 2 6= 0, the ∂y equation F(x,y) = 0 defines an implicit function y = f (x) in the neighbourhood of (0,1). Because the equation F(x,y) = x2 + y2 − 1 = 0 is simple, we can even derive the explicit expression of the implicit function to be p y = f (x) = 1 − x2.

 ∂F The following four examples will demonstrate the necessity of the assumption (x ,y ) 6= 0 in Theorem ∂y 0 0 6.1.1 to be able to conclude the existence of an implicit function. ∂F Example 6.2 The point (x ,y ) = (0,0) satisfies the equation F(x,y) = x2 + y2 = 0. Further, (x ,y ) =  0 0 ∂y 0 0 2y0 = 0. Therefore, the assumptions of Theorem 6.1.1 are not satisfied and we cannot conclude anything about the existence of an implicit function in a neighbourhood of (0,0). However, because the equation F(x,y) = x2 + y2 = 0 is satisfied only for the point (0,0), there does not exist an implicit function defined in the neighbourhood of (0,0). 

 Example 6.3 The equation

F(x,y) = x2 + 2y2 − 1 = 0 6.1 Implicit functions of one variable 67

2 ∂F defines an ellipse in R . The point (x0,y0) = (1,0) is such that F(1,0) = 0 and (x0,y0) = 4y0 = 0, thus the ∂y assumptions of Theorem 6.1.1 are not satisfied. The equation F(x,y) = 0 does not define an implicit function in a neighbourhood of the point (1,0) from two reasons: 1. There does not exist y ∈ R such that F(x,y) = 0 for x > 1. 2. For every x ∈ (1 − δ,1), where 0 < δ < 2, there are two values r r 1 − x2 1 − x2 y = and y = − 1 2 2 2

such that both (x,y1) and (x,y2) satisfy the equation F(x,y) = 0. Hence, in arbitrarily small neighbourhood q q 1−x2 1−x2 of (x0,y0) there are always two distinct points (x,y1 = 2 ) and (x,y2 = − 2 ) satisfying the equation F(x,y) = 0.



 Example 6.4 The set of all points satisfying the equation

F(x,y) = y2 − x2 = 0 is the set of points of the lines y = x and y = −x. For every x close to 0 there exist two values y1 = x and y2 = −x close to 0 such that F(x,y1,2) = 0. Therefore, the equation y2 − x2 = 0 does not define an implicit function in the neighbourhood of (0,0). Note ∂F that F(0,0) = 0 = (0,0). ∂y However, the equation y2 − x2 = 0 defines the implicit function y = x for instance in the neighbourhood ∂F of the point (x ,y ) = (2,2). Note that F(2,2) = 0 and (2,2) = 4 6= 0 which means the assumptions of 0 0 ∂y Theorem 6.1.1 are satisfied. 

 Example 6.5 Let us consider the equation

F(x,y) = y3 − x = 0.

Theorem 6.1.1 cannot be applied to decide the existence of an implicit function in the neighbourhood of (0,0) ∂F since F(0,0) = 0 and (0,0) = 0. However, the equation F(x,y) = 0 is equivalent to ∂y √ y = 3 x √ and thus the equation y3 − x = 0 defines the implicit function y = 3 x in the neighbourhood of (0,0) (even in 2 R ). 

If the equation F(x,y) = 0 defines an implicit function y = f (x) in the neighbourhood of (x0,y0), then certainly y0 = f (x0). In general, without further theory, nothing else could be said about the function values of f around x0. In the example below we will apply the Newton’s method to approximate values of an implicit function.

 Example 6.6 Let us consider the equation

x3 + y3 − 3xy − 3 = 0.

This equation is satisfied for example by the point (x0,y0) = (1,2). Let us denote

F(x,y) = x3 + y3 − 3xy − 3.

Then ∂F ∂F = 3y2 − 3x and (x ,y ) = 9 6= 0. ∂y ∂y 0 0 68 Chapter 6. Implicit functions

The given equation therefore implicitly defines a function y = f (x) in the neighbourhood of (x0,y0). We know that f (1) = 2. Further, let us assume f is defined on O0.11(1) = (0.89,1.11). What would be the function value of f at x = 0.9? By substituting x = 0.9 into the equation x3 + y3 − 3xy − 3 = 0 one derives

y3 − 2.7y − 2.271 = 0.

This equation has to have exactly one solution y = f (0.9) in the neighbourhood of the point y0 = 2. The first approximation of this solution by the Newton’s method is . y = f (0.9) = 1.963783.

 The theorem below shows us how to differentiate implicit functions. The computed derivatives can be used to decide monotonicity, convexity and concavity of implicit functions, to construct Taylor polynomials to approximate implicit functions, to determine their tangents and differentials.

2 k Theorem 6.1.2 Let G ⊆ R be an open set and let F ∈ C (G) for k ∈ N. Let (x0,y0) ∈ G be a point such that ∂F F(x ,y ) = 0 and (x ,y ) 6= 0. 0 0 ∂y 0 0

Then the derivative of an implicit function y = f (x) defined by the equation F(x,y) = 0 in a neighbourhood of (x0,y0) is given as

∂F (x, f (x)) f 0(x) = − ∂x for x ∈ (x − δ,x + δ) for some δ > 0. (6.1) ∂F 0 0 ∂y (x, f (x))

Proof. The proof follows from the rule on how to differentiate composed functions: The composed function h(x) = F(x, f (x)) is identically zero on a neighbourhood of x0 and thus

∂F ∂F dy ∂F ∂F 0 = h0(x) = (x, f (x)) · 1 + (x, f (x)) · = (x, f (x)) + (x, f (x)) f 0(x) ∂x ∂y dx ∂x ∂y

∂F ∂F ∂F in the same neighbourhood. Because ∂y (x0,y0) 6= 0 and because ∂y (x,y) is continuous, the derivative ∂y (x,y) 6= 0 in a neighbourhood of (x0,y0). Therefore,

∂F (x, f (x)) f 0(x) = − ∂x for x ∈ (x − δ,x + δ) for some δ > 0. ∂F 0 0 ∂y (x, f (x))



 Example 6.7 Let us differentiate the implicit function y = f (x) given by the equation

F(x,y) = x3 + y3 − 3xy − 3 = 0 in the neighbourhood of (x0,y0) = (1,2). According to the results in Example 6.6 the function f exists. Let us recall that

∂F ∂F (x,y) = 3y2 − 3x and (1,2) = 9 6= 0. ∂y ∂y

Further,

∂F ∂F (x,y) = 3x2 − 3y and (1,2) = −3. ∂x ∂x 6.1 Implicit functions of one variable 69

Then

∂F (1,2) −3 1 f 0(1) = − ∂x = − = . ∂F 9 3 ∂y (1,2) Moreover,

∂F (x, f (x)) f (x) − x2 f 0(x) = − ∂x = > 0 ∂F f (x)2 − x ∂y (x, f (x)) for x ∈ (1 − δ,1 + δ) for δ > 0. This means f is increasing in a neighbourhood of the point x0 = 1. 

R Higher order derivatives of an implicit function y = f (x) can be derived by differentiating the formula (6.1). For instance:

 2 2   2 2  ∂ F (x, f (x)) + ∂ F (x, f (x)) f 0(x) ∂F (x, f (x)) − ∂ F (x, f (x)) + ∂ F (x, f (x)) f 0(x) ∂F (x, f (x)) 00 ∂x2 ∂y∂x ∂y ∂x∂y ∂y2 ∂x f (x) = − 2  ∂F  ∂y (x, f (x))

or shortly

 2 2   2 2  ∂ F (x,y) + ∂ F (x,y)y0 ∂F (x,y) − ∂ F (x,y) + ∂ F (x,y)y0 ∂F (x,y) 00 ∂x2 ∂y∂x ∂y ∂x∂y ∂y2 ∂x y = − 2  ∂F  ∂y (x,y)

where y is viewed as the function y = f (x).

 Example 6.8 From Example 6.6 and 6.7, the equation

F(x,y) = x3 + y3 − 3xy − 3 = 0

0 1 defines an implicit function y = f (x) in the neighbourhood of (x0,y0) = (1,2) such that f (1) = 2 and f (1) = 3 . By differentiating the expression

∂F (x, f (x)) f (x) − x2 f 0(x) = − ∂x = ∂F f (x)2 − x ∂y (x, f (x)) one derives ( f 0(x) − 2x)( f (x)2 − x) − ( f (x) − x2)(2 f (x) f 0(x) − 1) f 00(x) = ( f (x)2 − x)2 and thus

( f 0(1) − 2)( f (1)2 − 1) − ( f (1) − 1)(2 f (1) f 0(1) − 1) (1 − 6) − (4 1 − 1) −5 − 1 −16 f 00(1) = = 3 = 3 = < 0 ( f (1)2 − 1)2 32 32 27 which means that f is concave in a neighbourhood of the point x0 = 1. Note that with the notation y = f (x) one may write

y − x2 y0 = y2 − x and by differentiating with respect to x:

(y0 − 2x)(y2 − x) − (y − x2)(2yy0 − 1) y00 = . (y2 − x)2

0 Consequently one substitutes the values x = 1, y = 2 and y = 1/3.  70 Chapter 6. Implicit functions

0 R Let F(x,y) = 0 define an implicit function y = f (x) in a neighbourhood of (x0,y0). One may compute f (x) by differentiating the equation F(x,y) = 0 with respect to x so that y is considered being a function of x. By repeated differentiation one derives higher order derivatives of f .

 Example 6.9 Consider the equation

F(x,y) = x3 + y3 − 3xy − 3 = 0 from Example 6.6 which defines an implicit function y = f (x) in the neighbourhood of (1,2). Then

∂F (x, f (x)) = 3x2 + 3 f 2(x) f 0(x) − 3 f (x) − 3x f 0(x) = 0 ∂x and therefore

3x2 − 3 f (x) f (x) − x2 1 f 0(x) = = and f 0(1) = . −3 f 2(x) + 3x f 2(x) − x 3

Because

∂ 2F (x, f (x)) = 6x + 6 f (x)( f 0(x))2 + 3 f 2(x) f 00(x) − 3 f 0(x) − 3 f 0(x) − 3x f 00(x) = 0, ∂x2

2(x + f (x)( f 0(x))2 − f 0(x)) 16 f 00(x) = and f 00(1) = − . − f 2(x) + x 27 From the derivatives one concludes that the implicit function f is increasing and concave in a neighbourhood of 1 as it was done in Example 6.7 and 6.8. (k) For the implicit function f one can compute f (1) for k ∈ N0 and thus one can approximate f by Taylor polynomial of degree k in a neighbourhood of 1. For instance,

. 1 1 8 f (x) = T (x) = f (1) + f 0(1)(x − 1) + f 00(1)(x − 1)2 = 2 + (x − 1) − (x − 1)2. 2 2! 3 27

∂F The point (1,−1) satisfies the equation F(x,y) = 0, i.e. F(1,−1) = 0. Because ∂y (1,−1) = 0 we do not know whether F(x,y) = 0 defines an implicit function y = f (x) in a neighbourhood of (1,−1). However, ∂F 2 because ∂x (1,−1) = 3 · 1 − 3 · (−1) = 6 6= 0, the equation F(x,y) = 0 defines an implicit function x = g(y) in a neighbourhood of (1,−1). The following holds for g: • g(−1) = 1, ∂F 2 2 0 ∂y 3y −3x y −x 0 • g (y) = − ∂F = − 3x2−3y = y−x2 and thus g (−1) = 0, ∂x 00 (2y−x0)(y−x2)−(y2−x)(1−2xx0) 00 • g (y) = (y−x2)2 and thus g (−1) = 1, which means that g has a local minimum at the point −1 and g is convex in a neighbourhood of −1. Now it is obvious the equation F(x,y) = 0 does not define an implicit function y = f (x) in a neighbourhood of (1,−1). 

 Example 6.10 Show that the equation

e2x + ey + x + 2y − 2 = 0 defines an implicit function y = f (x) in a neighbourhood of the point (0,0). Compute f 000(0). First let us denote the left-hand side of the equation by F(x,y). Because

F(0,0) = e0 + e0 + 0 + 0 − 2 = 0 and because  2x  ∂F 2e + 1 3 (0,0) = y = = 1 6= 0, ∂y e + 2 (x,y)=(0,0) 3 the equation F(x,y) = 0 implicitly defines a function y = f (x) in a neighbourhood of the point (0,0). Let us compute the third derivative of the function f at the point x0 = 0 in two ways: 6.1 Implicit functions of one variable 71

• By differentiating the formula for the first derivative: The first derivative of the function f (x) in a neighbourhood of 0 defined by the equation F(x,y) = 0 equals

∂F (x, f (x)) 2e2x + 1 f 0(x) = − ∂x = − . ∂F e f (x) + 2 ∂y (x, f (x)) Therefore, since f (0) = 0, f 0(0) = −1. Further,

4e2x(e f (x) + 2) − (2e2x + 1)e f (x) f 0(x) f 00(x) = [ f 0(x)]0 = − (e f (x) + 2)2

00 12+3 5 and, by evaluating the formula for x = 0, f (0) = − 32 = − 3 . Finally, 8e2x(e f (x) + 2) − (2e2x + 1)e f (x)( f 0(x))2 − (2e2x + 1)e f (x) f 00(x)
(e f (x) + 2)2 f 000(x) = [ f 00(x)]0 = − (e f (x) + 2)4 4e2x(e f (x) + 2) − (2e2x + 1)e f (x) f 0(x)
2(e f (x) + 2)e f (x) f 0(x) + (e f (x) + 2)4 and then (24 − 4 + 4 − 3 + 5)9 + (12 + 3)6 36 f 000(0) = − = − = −4. 34 9 • By differentiating the equation: Because the equation F(x,y) = 0 implicitly defines a function y = f (x) in a neighbourhood of (0,0), we substitute f (x) into y, where x is considered from a neighbourhood of 0:

e2x + e f (x) + x + 2 f (x) − 2 = 0.

By differentiating both sides of this equation with respect to x one obtains

2e2x + e f (x) f 0(x) + 1 + 2 f 0(x) = 0

which specifies the first derivative of f in a neighbourhood of 0. Specifically, f 0(0) = −1. By further differentiation of this equation, the relation specifying the second derivative of f is obtained:

4e2x + e f (x)( f 0(x))2 + e f (x) f 00(x) + 2 f 00(x) = 0.

00 5 00 Thus, the substitution x = 0 implies f (0) = − 3 . Lastly, differentiation of the equation specifying f (x) leads to the equation which specifies f 000(x) for x in a neighbourhood of 0:

8e2x + e f (x)( f 0(x))3 + e f (x)2 f 0(x) f 00(x) + e f (x) f 0(x) f 00(x) + e f (x) f 000(x) + 2 f 000(x) = 0.

Therefore, by substituting x = 0, f 000(0) = −4.



 Example 6.11 Show that the equation

ln(x + y) = x + y − xy − x2 − y2 defines an implicit function y = f (x) in a neighbourhood of the point (0,1). Determine the tangent line to the graph of f at the point (0,1). 2 2 ∂F Let F(x,y) = ln(x + y) − x − y + xy + x + y . Then, because F(0,1) = ln(1) − 1 + 1 = 0 and ∂y (0,1) =   1 − 1 + x + 2y = 2 6= 0, the equation F(x,y) = 0 implicitly defines a function y = f (x) in a x+y (x,y)=(0,1) neighbourhood of the point (0,1). Because

 1  ∂F − 1 + y + 2x (0,1) x+y (x,y)=( , ) f 0(0) = − ∂x = − 0 1 = −1/2, ∂F  1  ∂y (0,1) − 1 + x + 2y x+y (x,y)=(0,1) 72 Chapter 6. Implicit functions

the tangent line to the graph of f at the point (0,1) is given as x y = f (0) + f 0(0)(x − 0), i.e. y = 1 − . 2



 Example 6.12 Show that the equation ex − ey − x − y = 0 defines an implicit function y = f (x) in a neighbourhood of the point (0,0). Decide whether f is increasing, decreasing, convex, concave in a neighbourhood of the point 0. Approximate the function f by Taylor polynomial of degree 2 about the point 0. Let us denote F(x,y) = ex − ey − x − y. Because F(0,0) = e0 − e0 − 0 − 0 = 0, ∂F (0,0) = (−ey − 1)| = −2 6= 0, ∂y y=0 the equation F(x,y) = 0 implicitly defines a function y = f (x) in a neighbourhood of (0,0). To decide monotonicity and convexity/concavity of f ia a neighbourhood of 0 and to approximate f (x) by 0 00 T2(x) for x close to 0, we first compute f (0), f (0) and f (0).

f (0) = 0 ( because F(0,0) = 0 ) 0 ∂ x f (x) 0 0 f (0) = 0 ( because ∂x F(x, f (x)) = e − e f (x) − 1 − f (x) = 0 and thus, for x = 0, 1 − f 0(0) − 1 − f 0(0) = 0 which implies f 0(0) = 0 ) 2 f 00( ) = / ∂ F(x, f (x)) = ex − e f (x)( f 0(x))2 − e f (x) f 00(x) − f 00(x) = x = , 0 1 2 ( because ∂x2 0 and thus, for 0 1 − 2 f 00(0) = 0 which implies f 00(0) = 1/2 ) Hence, the function f has a local minimum at 0, it is convex in a neighbourhood of 0, it is decreasing in a left neighbourhood of 0 and increasing in a right neighbourhood of 0. Further, . f 00(0) 1 f (x) = T (x) = f (0) + f 0(0)(x − 0) + (x − 0)2 = x2 2 2 4 for x in a neighbourhood of 0.



6.1.1 Normal vector to a curve

The gradient ∇ f (x0,y0) of a function f (x,y) at the point (x0,y0) is a normal vector to the contour line of the graph of f corresponding to the cutting plane z = z0 passing through the point (x0,y0).

Figure 6.2: Geometrical meaning of ∇ f (x0,y0)

The points of the contour line are the points which satisfy the equation

F(x,y) = f (x,y) − z0 = 0. 6.2 Implicit functions of several variables 73

Theorem 6.1.3 Consider an equation F(x,y) = 0 and a point (x0,y0) such that F(x0,y0) = 0. Let ∇F(x0,y0) 6= (0,0). Then the points (x,y) ∈ O(x0,y0) which satisfy the equation F(x,y) = 0 form a curve which 1. passes through the point (x0,y0),  ∂F ∂F  2. has the normal vector ∇F(x0,y0) = ∂x (x0,y0), ∂y (x0,y0) at the point (x0,y0).

∂F ∂F Proof. Because ∇F(x0,y0) 6= (0,0), at least one of the derivatives ∂x (x0,y0) and ∂y (x0,y0) is non-zero. Let us ∂F assume without loss of generality that ∂y (x0,y0) 6= 0. Then the equation F(x,y) = 0 defines an implicit function y = f (x) in a neighbourhood of the point (x0,y0). Further,

∂F (x0,y0) f 0(x ) = − ∂x 0 ∂F ∂y (x0,y0)

which specifies the slope of the tangent line to the graph of f at the point (x0,y0). The directional vector of this tangent line is thus

 ∂F ∂F  d~ = − (x ,y ), (x ,y ) . ∂y 0 0 ∂x 0 0

Because the normal vector~n to the graph of f at the point (x0,y0) is perpendicular to the directional vector d~ of the tangent line to f at (x0,y0), one obtains ∂F ∂F  ~n = (x ,y ), (x ,y ) = ∇F(x ,y ) ∂x 0 0 ∂y 0 0 0 0 ~ (this can be derived from~n · d = 0). 

 Example 6.13 Consider the equation

ln(x + y) = x + y − xy − x2 − y2

defining an implicit function y = f (x) in a neighbourhood of the point (0,1) studied in Example 6.11. Let us determine the tangent line to the graph of f at the point (0,1) in another way. ∂F ∂F Because ∂x (0,1) = 1 and ∂y (0,1) = 2, the vector ∇F(0,1) = (1,2)

is the normal vector to the tangent line to the graph of f at the point (0,1). Therefore, the equation defining the tangent line is of the form

1x + 2y + c = 0,

where c is such that 0 + 2 · 1 + c = 0 holds since the point (x0,y0) has to lie on the tangent line. Hence, c = −2 and the tangent line is

x + 2y − 2 = 0.



6.2 Implicit functions of several variables Let us generalize the preceding material to the implicit functions of several variables.

Definition 6.2.1 An equation F(x1,...,xn,z) = 0 defines an implicit function z = f (x1,...,xn) on a neigh- 0 0 0 bourhood of a point x = (x1,...,xn,z0) if 0 0 1. F(x1,...,xn,z0) = 0 and 0 2. ∃δ > 0,ε > 0 such that ∀x = (x1,...,xn) ∈ Oδ (x )∃!z ∈ (z0 − ε,z0 + ε) : F(x1,...,xn,z) = 0. 74 Chapter 6. Implicit functions

0 0 Apparently, F(x1,...,xn, f (x1,...,xn)) = 0 for each (x1,...,xn) ∈ Oδ (x ) and therefore f (x ) = z0.

k n+1 0 0 Theorem 6.2.1 Let F ∈ C (G), where G ⊆ R is an open set, k ∈ N. Let (x1,...,xn,z0) be such a point that 0 0 1. F(x1,...,xn,z0) = 0, ∂F 0 0 2. ∂z (x1,...,xn,z0) 6= 0. Then the equation F(x1,...,xn,z) = 0 defines an implicit function z = f (x1,...,xn) on a neighbourhood of 0 0 k 0 0 the point (x1,...,xn,z0). Additionally, f ∈ C (Oδ ((x1,...,xn)) for a certain δ > 0.

Theorem 6.2.2 If the assumptions of the previous theorem are satisfied then the partial derivatives of the implicitly defined function z = f (x1,...,xn) are given by

∂F ∂ f (x1,...,xn, f (x1,...,xn)) (x ,...,x ) = − ∂xi , i = 1,...,n. (6.2) 1 n ∂F ∂xi ∂z (x1,...,xn, f (x1,...,xn))

R • The relation (6.2) can be written in the form

∂F ∂z = − ∂xi , i = ,...,n. F 1 ∂xi ∂ ∂z • Higher-order partial derivatives of f can be derived by respective differentiation of (6.2).

 Example 6.14 The equation

F(x,y,z) = x2 + y2 + xyz + xz2 = 0 defines an implicit function z = f (x,y) on a neighbourhood of the point (−1,1,1) since • F(−1,1,1) = 1 + 1 − 1 − 1 = 0 and ∂F • (−1,1,1) = (xy + 2xz)| = −1 − 2 = −3 6= 0. ∂z (x,y,z)=(−1,1,1) Further,

∂ f ∂F (x,y, f (x,y)) 2x + y f (x,y) + f (x,y)2 (x,y) = − ∂x = − ∂x ∂F xy + 2x f (x,y) ∂z (x,y, f (x,y)) and

∂F ∂ f (x,y, f (x,y)) 2y + x f (x,y) (x,y) = − ∂y = − . ∂y ∂F xy + 2x f (x,y) ∂z (x,y, f (x,y)) In particular,

∂ f 2(−1) + 1 + 1 ∂ f 2 − 1 1 (−1,1) = − = 0 and (−1,1) = − = . ∂x −1 − 2 ∂y −1 − 2 3



 Example 6.15 Let us determine the tangent plane to the surface given by the equation

xz3 + yz2 + xy + 1 = 0 at the point (2,−1,1). First, let us check the equation defines an implicit function z = f (x,y) for (x,y) ∈ Oδ (2,−1) for a certain δ > 0: • F(2,−1,1) = 2 − 1 − 2 + 1 = 0, 6.2 Implicit functions of several variables 75

∂F • (2,−1,1) = (3xz2 + 2yz)| = 6 − 2 = 4 6= 0. ∂z (x,y,z)=(2,−1,1) Because f (2,−1) = 1, ∂F !  3  ∂ f x (x,y, f (x,y)) f (x,y) + y 0 (2,−1) = − ∂ = − = − = 0, ∂x ∂F (x,y, f (x,y)) 3x f (x,y)2 + 2y f (x,y) 6 − 2 ∂z (x,y)=(2,−1) (x,y)=(2,−1) ∂F (x,y, f (x,y))!  2  ∂ f ∂y f (x,y) + x 1 + 2 3 (2,−1) = − = − = − = − , ∂y ∂F (x,y, f (x,y)) 3x f (x,y)2 + 2y f (x,y) 6 − 2 4 ∂z (x,y)=(2,−1) (x,y)=(2,−1) the tangent plane is of the form 3 z = 1 + 0(x − 2) − (y + 1) 4 or, equivalently, 3y + 4z − 1 = 0.



 Example 6.16 The equation ln(z) + xyz = 0 defines an implicit function z = f (x,y) on a neighbourhood of the point (−1,0,1) since for F(x,y,z) = ln(z) + xyz the following holds: • F(−1,0,1) = 0 − 0 = 0, ∂F 1  • (−1,0,1) = + xy | = 1 − 0 = 1 6= 0. ∂z z (x,y,z)=(−1,0,1) Let us approximate the function f (x,y) by a Taylor polynomial T2(x,y) of degree 2. For that we compute all partial derivatives of f at the point (−1,0) up to order 2: f (−1,0) = 1, ∂ f yz ∂ f (x,y) = − ⇒ (−1,0) = 0, x 1 x ∂ z + xy ∂ ∂ f xz ∂ f (x,y) = − ⇒ (−1,0) = 1, y 1 y ∂ z + xy ∂ !! ∂ 2 f ∂ y f (x,y) (−1,0) = − ∂x2 ∂x 1 + xy f (x,y) (x,y)=(−1,0)     − ∂ f (x,y)  ∂ f 1 ∂x y (x,y) + xy − y f (x,y) 2 + y  ∂x f (x,y) f (x,y)  =  2  = 0,   1   f (x,y) + xy (x,y)=(−1,0)   !! 2 f y f (x,y) −1 ∂ ∂   (−1,0) = − 1 = 2 = −1 ∂y∂x ∂y + xy  1   f (x,y) (x,y)=(−1,0) + xy f (x,y) (x,y)=(−1,0) !! ∂ 2 f ∂ x f (x,y) (−1,0) = − ∂y2 ∂y 1 + xy f (x,y) (x,y)=(−1,0)   ∂ f (x,y) 1
 −1 ∂ f (x,y)  −x ∂y z + xy + xz f 2(x,y) ∂y + x =   = 3.   2  1 + xy f (x,y) (x,y)=(−1,0) 76 Chapter 6. Implicit functions

Alternatively, one may compute the derivatives of f by differentiating the equality

ln( f (x,y)) + xy f (x,y) = 0.

By differentiating it once with respect to x one obtains

1 ∂ f (x,y) ∂ f (x,y) + y f (x,y) + xy = 0. (6.3) f (x,y) ∂x ∂x

By differentiating it once with respect to y one obtains

1 ∂ f (x,y) ∂ f (x,y) + x f (x,y) + xy = 0. (6.4) f (x,y) ∂y ∂y

By substituting x = −1, y = 0 and f (−1,0) = 1 into (6.3) and (6.4) one obtains

∂ f ∂ f (−1,0) = 0 and (−1,0) = 1. ∂x ∂y Further, the differentiation of (6.3) with respect to x leads to

1 ∂ f (x,y)2 1 ∂ 2 f (x,y) ∂ f (x,y) ∂ 2 f (x,y) − + + 2y + xy = 0 f 2(x,y) ∂x f (x,y) ∂x2 ∂x ∂x2

which implies, for x = −1 and y = 0, that

∂ 2 f (−1,0) = 0. ∂x2 The differentiation of (6.3) with respect to y leads to

1 ∂ f (x,y) ∂ f (x,y) 1 ∂ 2 f (x,y) ∂ f (x,y) ∂ f (x,y) ∂ 2 f (x,y) − + + f (x,y) + y + x + xy = 0 f 2(x,y) ∂y ∂x f (x,y) ∂y∂x ∂y ∂x ∂y∂x

which implies, for x = −1 and y = 0, that

∂ 2 f (−1,0) = −1. ∂y∂x The differentiation of (6.4) with respect to y leads to

1 ∂ f (x,y)2 1 ∂ 2 f (x,y) ∂ f (x,y) ∂ 2 f (x,y) − + + 2x + xy = 0 f 2(x,y) ∂y f (x,y) ∂y2 ∂y ∂y2

which implies, for x = −1 and y = 0, that

∂ 2 f (−1,0) = 3. ∂y2

Finally, there follows the approximation of f (x,y) by a Taylor polynomial of degree 2: . 1 f (x,y) = T (x,y) = 1 + 0(x + 1) + 1(y − 0) + 0(x + 1)2 + 2(−1)(x + 1)(y − 0) + 3(y − 0)2
2 2 3 = 1 + y − (x + 1)y + y2. 2



6.2.1 Normal vector to a plane 6.2 Implicit functions of several variables 77

Theorem 6.2.3 Consider an equation F(x,y,z) = 0 and a point (x0,y0,z0) such that F(x0,y0,z0) = 0. Let ∇F(x0,y0,z0) 6= (0,0,0). Then the points (x,y,z) from a neighbourhood of (x0,y0,z0) which satisfy the equation F(x,y,z) = 0 form a surface containing the point (x0,y0,z0) whose normal vector in that point is

∂F ∂F ∂F  ∇F(x ,y ,z ) = (x ,y ,z ), (x ,y ,z ), (x ,y ,z ) . 0 0 0 ∂x 0 0 0 ∂y 0 0 0 ∂z 0 0 0

∂F Proof. Let ∇F(x0,y0,z0) 6= (0,0,0). Without loss of generality, let ∂z (x0,y0,z0) 6= 0. Then the surface F(x,y,z) = 0 for (x,y,z) ∈ O((x0,y0,z0)) is the graph of an implicitly defined function z = f (x,y). The tangent plane τ to this surface at the point (x0,y0,z0) is then given as

τ : z − z0 = k1(x − x0) + k2(y − y0) and thus

τ : k1x + k2y − z = d, where

d = k1x0 + k2y0 + z0 ∂F ∂ f (x0,y0, f (x0,y0)) k = (x ,y ) = − ∂x 1 ∂x 0 0 ∂F ∂z (x0,y0, f (x0,y0)) ∂F ∂ f (x0,y0, f (x0,y0)) k = (x ,y ) = − ∂y . 2 ∂y 0 0 ∂F ∂z (x0,y0, f (x0,y0))

The vector~n = (k1,k2,−1) is a normal vector to the plane τ. However, every vector of the form k~n for k ∈ R\{0} is a normal vector to τ. For instance, for ∂F k = − (x ,y , f (x ,y )), ∂z 0 0 0 0 the normal vector of τ is ∂F ∂F ∂F  ∇F(x ,y ,z ) = (x ,y ,z ), (x ,y ,z ), (x ,y ,z ) . 0 0 0 ∂x 0 0 0 ∂y 0 0 0 ∂z 0 0 0 

 Example 6.17 Let us determine the tangent plane τ to the surface given by the equation zx2 + y2 − 2z3 = 0 at the point (1,1,1). Because for the function F(x,y,z) = zx2 + y2 − 2z3: F(1,1,1) = 0, ∂F (1,1,1) = (x2 − 6z)| = −5 6= 0, ∂z (x,y,z)=(1,1,1) the equation F(x,y,z) = 0 defines an implicit function z = f (x,y) on a neighbourhood of (1,1,1). Further, because ∂F (1,1,1) = (2xz)| = 2, ∂x (x,y,z)=(1,1,1) ∂F (1,1,1) = (2y)| = 2, ∂y (x,y,z)=(1,1,1) the tangent plane τ computed as the tangent plane to the graph of f at the point (1,1,1) is given by its normal vector ∇F(1,1,1) = (2,2,−5) as τ : 2x + 2y − 5z = d, where d = 2 · 1 + 2 · 1 − 5 · 1 = −1.

 78 Chapter 6. Implicit functions

6.3 Exercises 1. (a) Verify that the equation

xy − y3 − 1 = 0

defines an implicit function y = f (x) in the neighbourhood of the point (2,1). (b) Compute f 0(2), f 00(2), f 000(2). (c) Sketch the graph of f in the neighbourhood of x = 2. (d) Determine the tangent line to the graph of f at the point (2,1). (e) Determine the Taylor polynomial of degree 2 for the function f about the point x0 = 2. 2. Show that the equation

ln(x + y) = x + y − xy − x2 − y2

defines an implicit function y = f (x) in a neighbourhood of the point (1,0). Determine the tangent line to the graph of y = f (x) at the point (1,0). 3. Approximate the values f (0.95) and f (1.05) of the implicit function y = f (x) defined by the equation

x3 + y3 − 3xy − 3 = 0

in the neighbourhood of the point (1,2). (Hint: use the Newton’s method for the approximation.) 4. Consider the equation

sin(x) + sin(y) + xy = 0.

(a) Specify at least three distinct points (xi,yi), i = 1,2,3 such that the equation above defines an implicit function y = fi(x) in a neighbourhood of (xi,yi). (b) Decide whether fi is increasing, decreasing, convex, concave in a neighbourhood of (xi,yi). (c) Approximate the function y = fi(x) in a neighbourhood of xi by a Taylor polynomial of degree 2. Approximate the function value of fi at the point xi + 0.2. (d) Compute the value of the function f1 at the point x1 + 0.2 by Newton’s method with the precision −5 10 . Compare this approximation of f1(x1 + 0.2) with the approximation from the previous step. 5. Consider the equation E and the point A = (a1,a2) given as (i) E : y + sin(y) − x = 0 A = (0,0) (ii) E : exy − x2 + y3 = 0 A = (−1,0) (iii) E : x2y3 + y − 3 = 0 A = (0,3) (iv) E : x2 + y2 − ln(y) − 2 = 0 A = (1,1) (v) E : 2x2y + x − 1 − sin(y) = 0 A = (0,1) x+y (vi) E : e − xy + 2y = 0 A = (√1,−1√) (vii) E : x3 + y3 − 3xy = 0 A = ( 3 2, 3 4) (viii) E : yex−1 + xey − 1 = 0 A = (1,0) (a) Does the equation E define a function in x on a neighbourhood of A? (b) Let y = f (x) be a function which is implicitly defined by the equation E on the neighbourhood of A. 0 00 000 Compute y (a1), y (a1), y (a1). (c) Does f have an extremum at x = a1? (d) Is A an inflexion point of f ? (e) Compute the tangent line to the graph of f at the point A. (f) Determine the Taylor polynomial of degree 3 of f at x0 = a1. (g) Sketch the graph of f in the neighbourhood of A. π 6. Compute the tangent line to the curve ysin(x) − cos(x) + y − 2 = 0 at the point A = ( 2 ,1). 7. Determine whether the equation F(x,y,z) = 0 defines a function z = f (x,y) in a neighbourhood of the point A = (a1,a2,a3). Does the function f have a local extremum at (a1,a2)? Compute the differential of f at (a1,a2). Determine the tangent plane to the surface given by the equation F(x,y,z) = 0 at the point A. Compute the Taylor polynomial of degree 2 for the function f at the point (a1,a2). 6.4 Answers 79

(a) F(x,y,z) = x3 + y3 + z3 − z − 1, A = (1,0,1) (b) F(x,y,z) = z3 − 2xz + y, A = (1,1,1) 2 2 π (c) F(x,y,z) = y + z − 2yz + x − 1 + 2 − 2arctan(xz), A = (1,1,1).

6.4 Answers 1. (a) (b) (c) (d) (e) 2. 3. 4. (a) (b) (c) (d) 5. (a) (b) (c) (d) (e) (f) (g) 6. 7. (a) (b) (c)