Calculus: Analytic Geometry and Calculus, with Vectors

172 Functions, limits, derivatives

15 Supposing that y is a differentiable function of x for which

x2 + xy(x) + [y(x)]2 = 3,

apply our fundamental formulas for calculating derivatives to obtain the formula

2x + Y(x)

Y,(x) x + 2y(x)

Hint: Equate the derivatives with respect to x of the two members of the given

equation. Remark: This process, by which we start with an equation involving

y(x) and [without obtaining an explicit formula for y(x)] obtain an explicit formula

for y'(x), is called "implicit differentiation." To gain understanding of this

terminology, we can note that the formula y = x + 1 says explicitly that y is

x + 1 while the equation y - x - 1 = 0 only implies, and hence says implicitly,

that y is x + 1. It is sometimes said that the equation x2 + y2 = I determines y

implicitly, but the fact is that the equation does not determine y. Saying that "y

is either 1 - x2 or --/1 - x2" does not determine y any more than saying

"a blonde did it" determines the culprit in a whodonit.

16 Write the first displayed formula of the preceding problem in the form

x2 + xy + y2 = 3

and use the Leibniz notation for derivatives to obtain the formula

dy 2x + y

dx x+2y

Observe, however, that the calculation is illusory unless y is a differentiable func-

tion for which the given equation holds.

17 Clarify matters relating to the two preceding problems by showing that

y is a differentiable function satisfying the given equation if

Y(x) _

-x - 2 (4 - x2)

(-2 < x < 2)

and also if

Y(x)=-x+ 3(4-x2

2 (-2 < x < 2).

18 A graph of the equation

x2+xy+y2=3

appears in Figure 1.592. Find the equations of the tangents to this graph at the

two points for which x = 0. Be sure to obtain results that agree with Figure

1.592.

19 It is not a simple matter to "solve" the equation

(1) ya + y = x

for y. If y is a differentiable function of x for which the equation holds, however,