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Implicit and explicit functions

In mathematics, an implicit function is a function in which the dependent variable has not been given "explicitly" in terms of the independent variable. To give a function fexplicitly is to provide a prescription for determining the output value of the function y in terms of the input value x:

y = f(x).

By contrast, the function is implicit if the value of y is obtained from x by solving an equation of the form:

R(x,y) = 0.

That is, it is defined as the level set of a function in two variables: one variable or the other may determine the other, but one is not given an explicit formula for one in terms of the other.

Implicit functions can often be useful in situations where it is inconvenient to solve explicitly an equation of the form R(x,y) = 0 for y in terms of x. Even if it is possible to rearrange this equation to obtain y as an explicit function f(x), it may not be desirable to do so since the expression of f may be much more complicated than the expression of R. In other situations, the equation R(x,y) = 0 may fail to define a function at all, and rather defines a kind of multiple-valued function. Nevertheless, in many situations, it is still possible to work with implicit functions. Some techniques from calculus, such as differentiation, can be performed with relative ease using implicit differentiation.

The implicit function theorem provides a link between implicit and explicit functions. It states that if the equation R(x, y) = 0 satisfies some mild conditions on its partial derivatives, then one can in principle solve this equation for y, at least over some small interval. Geometrically, the graph defined by R(x,y) = 0 will overlap locally with the graph of a function y = f(x).

Various numerical methods exist for solving the equation R(x,y)=0 to find an approximation to the implicit function y. Many of these methods are iterative in that they produce successively better approximations, so that a prescribed accuracy can be achieved. Many of these iterative methods are based on some form of Newton's method.


Inverse functions

Implicit functions commonly arise as one way of describing the notion of an inverse function. If f is a function, then the inverse function of f is a solution of the equation

x=f(y) \implies y = f^{-1}(x)

for y in terms of x. Intuitively, an inverse function is obtained from f by interchanging the roles of the dependent and independent variables. Stated another way, the inverse function is the solution y of the equation

R(x,y) = x-f(y) = 0.


  1. The natural logarithmy = ln(x) is the solution of the equation x − ey = 0.
  2. The product log is an implicit function given by x − yey = 0.

Algebraic functions

An algebraic function is a solution y for an equation R(x,y) = 0 where R is a polynomial of two variables. Algebraic functions play an important role in mathematical analysis and algebraic geometry. A simple example of an algebraic function is given by the unit circle:


Solving for y gives


Note that there are two "branches" to the implicit function: one where the sign is positive and the other where it is negative.


Not every equation R(x, y) = 0 has a graph that is the graph of a function, the circle equation being one prominent example. Another example is an implicit function given by x − C(y) = 0 where C is a cubic polynomial having a "hump" in its graph. Thus, for an implicit function to be a true function it might be necessary to use just part of the graph. An implicit function can sometimes be successfully defined as a true function only after "zooming in" on some part of the x-axis and "cutting away" some unwanted function branches. A resulting formula may only then qualify as a legitimate explicit function.

The defining equation R = 0 can also have other pathologies. For example, the implicit equation x = 0 does not define a function at all; it is a vertical line. In order to avoid a problem like this, various constraints are frequently imposed on the allowable sorts of equations or on the domain. The implicit function theorem provides a uniform way of handling these sorts of pathologies.

Implicit differentiation

In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions.

As explained in the introduction, y can be given as a function of x implicitly rather than explicitly. When we have an equation R(x, y) = 0, we may be able to solve it for y and then differentiate. However, sometimes it is simpler to differentiate R(x, y) with respect to x and then solve for dy/dx.


1. Consider for example

y + x = -5 \,

This function normally can be manipulated by using From Yahoo Answers

Question:I have the equation e^(x*y^2*z^3) +x +2y +3z =0 Does this equation define a differentiable function z=f(x,y) around (0,-1,1)? And what about x=h(y,z) around this point, does this function differentiable?

Answers:Call the function F(x,y,z,)=0 dF/dz = e^(x*y^2*z^3)*(3xy^2z^2)+3 and at (0,-1,1) is = 3 not 0 so the partial derivatives dz/dx and dz/dy exist at (0,-1,1) and z=f(x,y) is differentiable at this point 2) dF/dx= e^(xy^2z^3)*( y^2z^3)+1 = 2 so, this function is also differentiable at the given point If the given function would be e^(xy^2z^3)-x+2y+3z h(y,z ) would not be differentiable at(0,-1,1)

Question:Given the equation y^3 + 3x^2y + 13 = 0, how would it be possible to find the minimum y-coordinate of this equation using implicit differentiation? I got that dy/dx = (-2xy)/(y^2 + x^2) but have no clue where to go from there..

Answers:Use implicit differentiation to obtain an expression for dy/dx. A relative extrema (min/max/saddle) of the original function occurs wherever dy/dx = 0; so find the zeros of dy/dx (the values of {x,y} at which dy/dx = 0.) (Cheapest method -- avoid having to compute the 2nd derivative) For each zero of dy/dx, evaluate {y,dy/dx} at a value of x just to the right of the zero, and at a value of x just to the left of the zero. Let L = the value of y just to the left of the zero Let R = the value of y just to the right of the zero Let C = the value of y at the zero. If (C < L) and (C < R), then C is a relative minimum If (C > L) and (C < R), then C is a relative maximum Otherwise, C is a saddle point. Next, compare all the relative minimum values you found and select the one with the smallest value. Finally, consider the limit of the function as x goes to +infinity, and the limit as x goes to -infinity. Compare the value of these limits with the minimum relative minimum value you already found. Among those three values, choose the one closest to -infinity.

Question:I've read a whole bunch of tutorials on implicit differentiation and I still don't get it. There is an example in my book of: 2x^3 + 2y^3 - 9xy = 0 I don't understand why the 2's are differentiated but the 9 is left alone...Why aren't you supposed to take the derivative of the 9 as well? I'm just really confused about implicit differentiation as whole so any help would be great. Thanks!

Answers:2x^3 + 2y^3 - 9xy = 0 But the 9 is not left alone. It is treated as (9x)*y and then the product rule is applied to it. 6x^2 is obvious 6y^2dy/dx is also obvious I'll save the minus sign for a while d((9x)*y) / dx = 9x * dy/dx + 9y * dx/dx and dx/dx is 1, so this is 9x*dy/dx + 9y Thus the total equation reads 6x^2 + 6y^2 dy/dx - 9x*dy/dx - 9y = 0 Grouping the dy/dx terms to the left hand side of this equation and the others to the right, we get: 6y^2 * dy/dx - 9x*dy/dx = 9y - 6x^2 Factoring we now have: dy/dx * (6y^2 - 9x) = 9y - 6x^2 dy/dx = (9y - 6x^2) / ( 6y^2 - 9x) = [(3y - 2x^2) / (2y^2 - 3x)] ...<<<... Answer .

Question:Hi, does anyone know about implicit differentiation ? I was doing homework and I cant figure a couple problems : 1) implicit diff to find dy/dx : e^3x=sin(x+2y) 2) use implicit diff to find dy/dx and d^2y/dx^2 : x^2-y^3=8 well for problem #2 i got dy/dx but i dont know how to get d^2/dx^2. Does anyone know how to solve this? thanks.

Answers:Take derivatives of both sides then solve for dy/dx. For number 2, do it twice. Number 1: e^3x = sin (x + 2y) derivative of (3x) = derivative of (sin (x + 2y)) e^3x * 3 * dx/dx = cos (x + 2y) * (1*dx/dx + 2*dy/dx) Now solve for dy/dx. 3e^3x = cos (x + 2y) + 2cos (x + 2y) * dy/dx 3e^3x - cos (x + 2y) = 2cos (x + 2y) * dy/dx (3e^3x - cos (x + 2y)) / 2cos (x + 2y) = dy/dx dy/dx = (3e^3x - cos (x + 2y)) / 2cos (x + 2y) Number 2. x^2 - y^3 = 8 derivative of (x^2 - y^3) = derivative of (8) 2x*dx/dx - 3y^2 * dy/dx = 0 Solve for dy/dx. 2x - 3y^2 * dy/dx = 0 -3y^2 * dy/dx = -2x dy/dx = 2x / 3y^2 That's dy/dx. To find d^2 y / dx^2 we take the derivatives again. derivative of (dy/dx) = derivative of (2x / 3y^2) Use quotient rule. d^2 y / dx^2 = [3y^2 * (2*dx/dx) - 2x * (6y*dy/dx)] / (3y^2)^2 = [6y^2 - 12xy(dy/dx)] / 9y^4 Now plug in 2x/3y^2 for dy/dx and then simplify. d^2 y/ dx^2 = [6y^2 - 12xy(2x/3y^2)] / 9y^4 = [6y^2 - 8x^2/y] / 9y^4 = [6y^2(y/y) - 8x^2/y] / 9y^4 = [(6y^3 - 8x^2) / y] / 9y^4 d^2 y / dx^2 = (6y^3 - 8x^2) / 9y^5

From Youtube

Implicit Differentiation :Taking the derivative when y is defined implicitly.

Implicit Differentiation (part 2) :A hairier implicit differentiation problem.