Implicit Functions

Excerpts

• Implicit Functions

  In this case $y$ is said to be *implicitly* a function of $x$, or an implicit function.

• Implicit Functions

  For example, in $x^{2} - xy + y^{2} = a$, when $x$ is known, $y$ must be determined by the solution of an equation of the second degree.

• Implicit Functions

  Here, though we know that $y$ must be a function of $x$, we do not know, without further investigation, what function it is.

• Implicit Functions

  $y$ and $x$ are no longer independent; for, one of them being given, the other must be so taken that the equation $\phi(x, y) = 0$ may be satisfied.

• Implicit Functions

  Hence $\dfrac{dy}{dx}$ (meaning the limit) is $-\dfrac{1}{x^{2}}$, which will also be the result of (3) if $1 + \dfrac{1}{x}$ be substituted for $y$.

Equations

• Implicit Functions

  \phi(x, y) = 0

  The general implicit form of a relation between x and y, obtained by bringing all terms to one side of the equation.

• Implicit Functions

  x^{2} - xy + y^{2} = a

  An example equation in which y is implicitly, not explicitly, a function of x; for given x, y is found by solving a quadratic.

• Implicit Functions

  xy - x = 1

  A worked example relation between x and y, from which dy/dx is found by implicit differentiation and checked against the explicit solution.

• Implicit Functions

  u = \phi(x, y)

  Defines the auxiliary quantity u as a function of the two variables x and y, used to derive the rule for dy/dx.

• Implicit Functions

  du = \ux\, dx + \uy\, dy + \etc.

  The differential of u expressed through the partial derivatives of u with respect to x and y, with dx and dy as the small changes in x and y.

• Implicit Functions

  \ux\, dx + \uy\, dy = 0

  Since u stays zero along the curve, the small changes dx and dy satisfy this linear relation.

• Implicit Functions

  \frac{dy}{dx} = -\frac{\ux}{\uy}

  The derivative of an implicit function y of x is minus the ratio of the partial derivatives of phi with respect to x and y.

• Implicit Functions

  \frac{dy}{dx} = -\frac{\;\dfrac{du}{dx}\;}{\dfrac{du}{dy}}

  Equation (1): the derivative of y with respect to x equals minus the partial derivative of u with respect to x divided by the partial derivative of u with respect to y, valid when u is held at zero.

• Implicit Functions

  \dfrac{du}{dx} = y - 1

  For u = xy - x - 1, the partial derivative of u with respect to x is y - 1.

• Implicit Functions

  \dfrac{du}{dy} = x

  For u = xy - x - 1, the partial derivative of u with respect to y is x.

• Implicit Functions

  xy - x - 1 = 0

  The example relation written in the form phi(x, y) = 0 with all terms on one side.

• Implicit Functions

  \frac{dy}{dx} = -\frac{y - 1}{x}

  Equation (3): the derivative of y with respect to x for the example xy - x = 1, expressed in x and y.

• Implicit Functions

  y = 1 + \dfrac{1}{x}

  Solving xy - x = 1 for y gives y explicitly as a function of x.

• Implicit Functions

  -\dfrac{1}{x^{2}}

  The limit of the ratio dy/dx for the example is minus one over x squared, which agrees with equation (3) after substituting y = 1 + 1/x.

• Implicit Functions

  u = \Chg{\phi(x)}{\phi x}

  Defines u as a function of x alone, the first of two equations linking u, x and y.

• Implicit Functions

  u = \psi y

  Defines u as a function of y alone, the second of two equations linking u, x and y.

• Implicit Functions

  \phi x = \psi y

  The third equation implied by the two equations u = phi x and u = psi y, relating x and y.

• Implicit Functions

  du = \psi(y + dy) - \psi y

  The change in u when y changes by dy, computed from the second equation u = psi y.

• Implicit Functions

  \phi' x\, dx + \etc. = \psi' y\, dy + \etc.

  Equating the two expressions for du gives a relation between dx and dy, the etc. terms vanishing in the limit.

• Implicit Functions

  \frac{dy}{dx} = \frac{\phi' x}{\psi' y} = \frac{\;\dfrac{du}{dx}\;}{\dfrac{du}{dy}}

  Equation (2): dy/dx equals the ratio of the derivatives of phi and psi, which is also the ratio of the partial derivatives of u, in accordance with common algebra.

Problems

No exercises in this chapter.