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Encyclopedia > Implicit differentiation
Topics in Calculus

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In mathematics, to give an implicit function f is to give the graph of a function, as a relation. That is, one produces somehow a binary relation R(x,y) in the hope that it suffices to define y as a conventional function of x. Of course when there is a direct way to re-arrange the formula, making y the subject, this is straightforward.

Contents

Caveats

This may be true, as in the case of a graph that is a line; it may be true with some limitations, such as specifying that one cannot give a vertical line as a graph; it may be true with some limitations on the function domain, as when the relation is x = C(y) with C a cubic polynomial with a 'hump' in its graph; or it may be true only after also cutting R down to size, as in the case x = y2. That is, an implicit function can sometimes be defined successfully only by modifying the relation by 'zooming in' to some part of the x-axis, and 'cutting back' unwanted function branches. A resulting formula may qualify as an ordinary explicit function.


Implicit differentiation

In calculus, implicit differentiation can be applied to implicit functions. This is by an application of the chain rule, to calculate derivatives dy/dx without necessarily making y an explicit function of x.


Examples

Consider for example

y + x = -4.

This function can be differentiated normally by using algebra to change this equation to an explicit function:

f(x) = y = −x - 4;

such differentiation would result in a value of −1. Equally, one can use implicit differentiation;

dy/dx + dx/dx = 0 = dy/dx + 1; dy/dx = -1.

An example of an implicit function, for which implicit differentiation might be easier than attempting to use explicit differentiation, is

x4 + 2y2 = 8.

In order to differentiate this explicitly , one would have to obtain (via algebra)

f(x) = y = \pm\sqrt{\frac{8 - x^4}{2}},

and then differentiate this function. This creates two derivatives, one for y > 0 and another for y < 0. Implicit differentiation avoids this.


One might find it substantially easier to implicitly differentiate the implicit function;

4x3 + 4y(dy/dx) = 0;

thus,

dy/dx = −4x3 / 4y = −x3 / y.

Implicit function theorem

It can be shown that if R(x,y) is given by a smooth submanifold M in R2, and (a,b) is a point of this submanifold such that the tangent space there is not vertical, then M in some small enough neighbourhood of (a,b) is given by a parametrization (x,f(x)) where f is a smooth function. In less technical language, implicit functions exist and can be differentiated, unless the tangent to the supposed graph would be vertical. In the standard case where we are given an equation

F(x,y) = 0

the condition on F can be checked by means of partial derivatives.


For the important generalisation to functions of several variables, see implicit function theorem.


  Results from FactBites:
 
Implicit function Summary (931 words)
Implicit differentiation is a technique used in calculus to compute the derivative of functions even when an explicit formula for the function is unknown.
In general, the implicit function theorem states that an equation g(x, y) = C(where g is a continuously differentiable function and C is any constant) locally defines y as an implicit function of x in a neighborhood of any point where the partial derivative of g with respect to y is nonzero.
Sometimes standard explicit differentiation cannot be used, and in order to obtain the derivative, another method such as implicit differentiation must be employed.
Derivative - Wikipedia, the free encyclopedia (2214 words)
Implicit differentiation: If f(x,y) = 0 is an implicit function, we have: dy/dx = - (∂f / ∂x) / (∂f / ∂y).
Perhaps the most natural situation is that of functions between differentiable manifolds; the derivative at a certain point then becomes a linear transformation between the corresponding tangent spaces and the derivative function becomes a map between the tangent bundles.
For complex functions of a complex variable differentiability is a much stronger condition than that the real and imaginary part of the function are differentiable with respect to the real and imaginary part of the argument.
  More results at FactBites »

 

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