Implicit Functions

From the equation $3x + 2y + 1 = 0$ , we can find $y$ in terms of $x$ . In other words, we can find $\displaystyle{y = -\frac{1 - 3x}{2}}$ . In general, given a quadratic function $f(x,y)$ , if $y = g(x)$ always satisfies $f(x,g(x)) = 0$ , then we say the equation is an implicit function determined by the equation $f(x,y) = 0$ .

Understanding
Finding the implicit function determined by the equation is the same as solving the equation for . But for some type of and some , there may not be a satisfying the equation .

Theorem 4..8 Suppose that $f(x,y)$

is the class $C^{1}$ at $\boldsymbol{x}_{0} = (x_{0},y_{0})$ in the region $D$

. If

$\displaystyle f(x_{0},y_{0}) = 0, f_{y}(x_{0},y_{0}) \neq 0$

, then there is an unique implicit function $y = g(x)$

determined by $f(x,y) = 0$

in the neighborhood of $x = x_{0}$ which satisfies

	$\displaystyle 1.$	$\displaystyle f(x,g(x)) = 0$
	$\displaystyle 2.$	$\displaystyle g(x_{0}) = y_{0}$
	$\displaystyle 3.$	$\displaystyle g(x) {\rm is} C^{1}, \frac{dy}{dx} = -\frac{f_{x}(x,y)}{f_{y}(x,y)}$

$f_{x}(x,y)$ is a funciton of $x,y$ and $y$ is a function of $x$ . Thus we have the following figure.

**Figure 4.29:** Check
$\includegraphics[width=3.5cm]{SOFTFIG-4/implicitfct.eps}$

Thus, $\frac{df_{x}(x,y)}{dx} = f_{xx}(x,y) + f_{xy}(x,y)\frac{dy}{dx}$

NOTE In the neighborhood of the point satisfying $f_{y}(x,y) \neq 0$ , the implicit function $y$ exists and the implcit function is differentiable. Thus the total diffetential of $f$ is

$\displaystyle f_{x}dx + f_{y}dy = 0$

Now using the condition $f_{y} \neq 0$ ,

$\displaystyle \frac{dy}{dx} = -\frac{f_{x}}{f_{y}}$

This is 3. Furthermore, differentiate the above equation with respect to $x$

, we have

$\displaystyle \frac{d^{2}y}{dx^2} = -\frac{\frac{df_{x}}{dx}f_{y} - f_{x}\frac{df_{y}}{dx}}{f_{y}^{2}}$

Now we use

$\displaystyle \frac{df_{x}}{dx} = f_{xx} + f_{xy}\frac{dy}{dx} = f_{xx} - f_{xy}\frac{f_{x}}{f_{y}}$

$\displaystyle \frac{df_{y}}{dx} = f_{yx} + f_{yy}\frac{dy}{dx} = f_{xy} - f_{yy}\frac{f_{x}}{f_{y}}$

to obtain

$\displaystyle \frac{d^{2}y}{dx^2}$	$\displaystyle =$	$\displaystyle -\frac{(f_{xx} - f_{xy}\frac{f_{x}}{f_{y}})f_{y} - f_{x}(f_{xy} - f_{yy}\frac{f_{x}}{f_{y}})}{f_{y}^{2}}$
	$\displaystyle =$	$\displaystyle -\frac{f_{xx}f_{y}^{2} - 2f_{xy}f_{x}f_{y} + f_{yy}f_{x}^{2}}{f_{y}^{3}}$

Implicit Functions
$\frac{d^{2}y}{dx^2} = -\frac{f_{xx}f_{y}^{2} - 2f_{xy}f_{x}f_{y} + f_{yy}f_{x}^{2}}{f_{y}^{3}}$ is called 2nd order derivative of implicit function

Example 4..19 Find the derivative of a implicit function $y$

such as $\displaystyle{\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$ deteremined by

$\displaystyle x^{2} + y^{2} = 3xy$

SOLUTION Set $f(x,y) = x^{2} + y^{2} - 3xy$ and take total differential of $f(x,y)$ ,

$\displaystyle df = f_{x}dx + f_{y}dy = (2x - 3y)dx + (2y - 3x)dy = 0$

Thus,

$\displaystyle \frac{dy}{dx} = -\frac{2x-3y}{2y-3x}.$

Since $f_{xx} = 2, f_{xy} = -3, f_{yy} = 2$ ,

$\displaystyle \frac{d^{2}y}{dx^{2}}$	$\displaystyle =$	$\displaystyle -\frac{2(2y-3x)^{2} + 6(2x-3y)(2y-3x) + 2(2x-3y)^{2}}{(2y-3x)^{3}}$
	$\displaystyle =$	$\displaystyle \frac{10(x^2 - 3xy + y^2)}{(2y - 3x)^{2}} = 0\ensuremath{ \blacksquare}$

A problem of finding a differentiation of an implicit function is solved by taking total differentiatial

Check

Exercise 4..19 Find the derivative of a implicit function $y$

such as $\displaystyle{\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$ deteremined by

$\displaystyle 2x^2 + 5xy - 3y^2 = 1$

Exercise4-19
Since and .

SOLUTION Set $f(x,y) = 2x^2 + 5xy - 3y^2 - 1 = 0$ and find total differential of $f(x,y)$ . Then

$\displaystyle df = f_{x}dx + f_{y}dy = (4x + 5y)dx + (5x - 6y)dy = 0.$

Thus,

$\displaystyle \frac{dy}{dx} = -\frac{4x + 5y}{5x - 6y}.$

$\displaystyle \frac{d^2 y}{dx^2}$	$\displaystyle =$	$\displaystyle -\frac{(4 + 5\frac{dy}{dx})(5x - 6y) - (4x + 5y)(5 - 6\frac{dy}{dx})}{(5x - 6y)^2}$
	$\displaystyle =$	$\displaystyle -\frac{4(5x-6y)^2 - 10(4x+5y)(5x-6y) - 6(4x+5y)^2}{(5x - 6y)^3} = \frac{98}{(5x-6y)^3}\ensuremath{ \blacksquare}$

Example 4..20 Find $\displaystyle{z_{x}, z_{y}}$ for the implicit function $z = g(x,y)$

determined by the equation $xy + yz + zx = 1$

By writing $z = \frac{1 - xy}{y+x}$ , we can find $z_x = \frac{-y(y+x)-(1-xy)}{(y+x)^2} = -\frac{y^2 + 1}{(y+x)^2} = -\frac{y+z}{y+x}$ .

SOLUTION For $f(x,y,z) = xy + yz + zx - 1 = 0$ , set $x = t, y = s$ . Then,

$\displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\cdo... ...l f}{\partial z} \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} = 0$

Thus,

$\displaystyle \frac{\partial z}{\partial x} = -\frac{\frac{\partial f}{\partial... ...\frac{\partial f}{\partial z}\frac{\partial x}{\partial t}} = - \frac{y+z}{y+x}$

Similarly,

$\displaystyle \frac{\partial z}{\partial y} = -\frac{\frac{\partial f}{\partial... ...}\frac{\partial y}{\partial t}} = - \frac{x+z}{y+x} \ensuremath{ \blacksquare}$

Exercise 4..20 Find $\displaystyle{z_{x}, z_{y}}$ for the implicit function $z = g(x,y)$

determined by the equation $x^2 + y^2 + z^2 = 4$

SOLUTION For $f(x,y,z) = x^2 + y^2 + z^2 - 4$ , set $x = t, y = s$ .

$\displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\fra... ...l f}{\partial z}\frac{\partial z}{\partial x} \frac{\partial x}{\partial t} = 0$

Thus,

$\displaystyle \frac{\partial z}{\partial x} = -\frac{\frac{\partial f}{\partial... ...l f}{\partial z}\frac{\partial x}{\partial t}} = - \frac{2x}{2z} = -\frac{x}{z}$

Similarly,

$\displaystyle \frac{\partial z}{\partial y} = -\frac{\frac{\partial f}{\partial... ...al y}{\partial t}} = - \frac{2y}{2z} = -\frac{y}{z} \ensuremath{ \blacksquare}$

Theorem 4..9 If an implicit function $y,z$

is determined by the equations $f(x,y,z) = 0, g(x,y,z) = 0$

, then,

$\displaystyle \frac{dy}{dx} = - \frac{f_x g_z - f_z g_x}{f_y g_z - f_zg_y}.$

Theorem4-9
To find $\frac{dy}{dx}$ , we first find and then eliminate .

Example 4..21 Find $\displaystyle{\frac{dy}{dx}, \frac{dz}{dx}}$ for the implicit functions $y,z$

determined by the equations $\displaystyle{x^2 + y^2 + z^2 = 1, 2x + 3y - z = 1}$ .

SOLUTION Let $f(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 , g(x,y,z) = 2x + 3y - z - 1 = 0$ . Then totally differentiate $f$ and $g$ to get

$\displaystyle df = 2xdx + 2ydy + 2zdz = 0, dg = 2dx + 3dy - dz = 0$

Now delete $dz$

to get

$\displaystyle (2x+4z)dx + (2y+6z)dy = 0$

Check
From the equations , we delete by adding to the first equation, .

Thus,

$\displaystyle \frac{dy}{dx} = -\frac{2x+4z}{2y+6z}$

Similarly, delete $dy$

to get

$\displaystyle (6x-4y)dx + (6z + 2y)dz = 0$

Thus,

$\displaystyle \frac{dz}{dx} = -\frac{6x-4y}{6z+2y}\ensuremath{ \blacksquare}$

Exercise 4..21 Find $\displaystyle{\frac{dy}{dx}, \frac{dz}{dx}}$ for the implicit functions $y,z$

determined by the equations $\displaystyle{xyz = 1, xy + yz + zx = 1}$ .

Check
. Eliminate by multiplying to the former equation and multiplyimg to the latter equation. . Simplifying,

SOLUTION Let $f(x,y,z) = xyz -1 = 0, g(x,y,z) = xy + yz + zx - 1 = 0$ . Then find total derivatives.

$\displaystyle df$	$\displaystyle =$	$\displaystyle f_{x}dx + f_{y}dy + f_{z}dz = yzdx + xzdy + xydz = 0$	(4.3)
$\displaystyle dg$	$\displaystyle =$	$\displaystyle g_{x}dx + g_{y}dy + g_{z}dz = (y + z)dx + (x + z) dy + (y + x)dz = 0$	(4.4)

Now using the equation 4.3 and the equation 4.4 to eliminate $dz$

$\displaystyle ((y+x)yz - xy(y+z))dx + ((y+x)xz - xy(x+z))dy = 0$

Then

$\displaystyle \frac{dy}{dx} = -\frac{y^2 z - xy^2}{x^2 z - x^2 y} = \frac{y^2 (x-z)}{x^2 (z-y)}.$

This time, using the equation 4.3 and the equation 4.4 to eliminate $dy$

$\displaystyle ((x+z)yz - xz(y+z))dx + ((x+z)xy - xz(y+x))dz = 0$

$\displaystyle z^2(y-x)dx + x^2(y-z)dz = 0$

Thus

$\displaystyle \frac{dz}{dx} = \frac{z^2 (x-y)}{x^2 (y-z)}\ensuremath{ \blacksquare}$

Exercise A

1.

Find the $\displaystyle{\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$ for the implicit functions $y$

determined by the following functions．

(a) $\displaystyle{x - y^{2} = 1}$ (b) $\displaystyle{x^{2} + xy + 2y^{2} = 1}$ (c) $\displaystyle{x - e^{y} = 0}$

(d) $\displaystyle{x^{3} - 3xy + y^{3} = 1}$

2.

Find the $\displaystyle{\frac{dy}{dx}, \frac{dz}{dx}}$ for the implicit functions $y,z$

determined by the following functions．

(a) $\displaystyle{x^2 +y^2 + z^2 = 4, x + y + z = 1}$ (b) $\displaystyle{xyz = 1, x + y + z = 1}$

Exercise B

1.

Find $\displaystyle{\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$ for the implicit function $y$

determined by the following equations.

(a) $\displaystyle{2x^2+ 5xy - 3y^2 = 1}$ (b) $\displaystyle{y = e^{x+y}}$ (c) $\displaystyle{x^2 - y^2 = xy}$

(d) $\displaystyle{\log{\sqrt{x^2 + y^2}} = \tan^{-1}{\frac{y}{x}}}$

2.

Find $\displaystyle{\frac{dy}{dx}, \frac{dz}{dx}}$ for the implicit function $y,z$

determined by the following equations.

(a) $\displaystyle{x^2 +y^2 + z^2 = 4, x^2 + y^2 = 4x}$ (b) $\displaystyle{xyz = 1, xy + yz + zx = 1}$

3.

Find the equation of the tangent line and the normal line to the curve $2x^2 + 5y^2 = 12$

at the point $(1,\sqrt{2})$ .

4.

Find the equation of the tangent plane to the surface $\displaystyle{z = \tan^{-1}\frac{y}{x}}$ at the point $(1,1,\frac{\pi}{2})$ ．Find the equation of the normal line through $(1,1,\frac{\pi}{2})$ .

5.

Find the local extrema of $y = g(x)$

implicitly defined by the following equations．

(a) $\displaystyle{8x^2 +4xy + 5y^2 = 36}$ (b) $\displaystyle{x^{2}y + x + y = 0}$