- #1
- 19,851
- 10,882
Definition/Summary
The definition of a function y of x is explicit if it is an equation in which y appears only once, and on its own (usually by starting "y =").
In any other case, the definition of a function y of x is implicit.
Implicit differentiation of y with respect to x is a slightly misleading name for ordinary differentiation of the defining equation of y.
Therefore, it generally involves \frac{dy}{dx} more than once, or functions of y, and application of the chain rule:
\frac{df(y)}{dx}\,=\,f'(y) \frac{dy}{dx} .
Equations
x^2\,+\,y^2\,=\,1 is an implicit definition of y.
Its implicit derivative with respect to x is:
2x\,+\,2y\frac{dy}{dx}\,=\,0
(where the chain rule has been applied by differentiating y^2 with respect to y, and then multiplying by \frac{dy}{dx})
which in this case can be simplified to:
\frac{dy}{dx}\,=\,-\frac{x}{y}
Extended explanation
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
The definition of a function y of x is explicit if it is an equation in which y appears only once, and on its own (usually by starting "y =").
In any other case, the definition of a function y of x is implicit.
Implicit differentiation of y with respect to x is a slightly misleading name for ordinary differentiation of the defining equation of y.
Therefore, it generally involves \frac{dy}{dx} more than once, or functions of y, and application of the chain rule:
\frac{df(y)}{dx}\,=\,f'(y) \frac{dy}{dx} .
Equations
x^2\,+\,y^2\,=\,1 is an implicit definition of y.
Its implicit derivative with respect to x is:
2x\,+\,2y\frac{dy}{dx}\,=\,0
(where the chain rule has been applied by differentiating y^2 with respect to y, and then multiplying by \frac{dy}{dx})
which in this case can be simplified to:
\frac{dy}{dx}\,=\,-\frac{x}{y}
Extended explanation
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!