Implicit differentiation

Relation with the implicit function theorem

Intuitively, in order to be valid at a point ${\displaystyle (x_0,y_0)}$ implicit differentiation in the basic formulation requires that the function ${\displaystyle F}$ must genuinely depend on ${\displaystyle y}$ near the point ${\displaystyle (x_0,y_0)}$, for otherwise the equation ${\displaystyle F(x,y)=F(x_0,y_0)}$ cannot be solved for ${\displaystyle y}$ as a function of ${\displaystyle x}$ at all.

More is actually required: it is clear from the formula that the partial derivative ${\displaystyle F_y(x_0,y_0)}$ must be non-zero, because it appears in the denominator of the right-hand side of the formula for ${\displaystyle dy/d{x}_{(x_0,y_0)}}$. While this fact allows one to compute a numerical value, it is not obvious that the numerical value has the advertised meaning: namely, that it be the derivative of a function ${\displaystyle y(x)}$ whose graph is contained on the curve ${\displaystyle F(x,y)=0}$ through the point.

The implicit function theorem supplies the missing justification. It asserts as follows: suppose that ${\displaystyle F(x,y)}$ is a function such that both partial derivatives exist in a disc around ${\displaystyle (x,y)=(x_0,y_0)}$, and are continuous throughout the disc. Then if ${\displaystyle F_y(x_0,y_0)\ne 0}$, there is a differentiable function ${\displaystyle y(x)}$ defined in an open interval containing ${\displaystyle x=x_0}$, such that ${\displaystyle F(x,y(x))=0}$ throughout the interval, and whose derivative is given by the implicit differentiation formula at ${\displaystyle (x_0,y_0)}$. The function is, moreover, unique on a sufficiently small interval around ${\displaystyle x_0}$.

Examples

Example 1

Consider

$${\displaystyle y+x+5=0.}$$

This equation is easy to solve for y, giving

$${\displaystyle y=-x-5,}$$

where the right side is the explicit form of the function y(x). Differentiation then gives ⁠dy/dx⁠ = −1.

Alternatively, one can differentiate the original equation:

$${\displaystyle \begin{array}{cc}\frac{dy}{dx}+\frac{dx}{dx}+\frac{d}{dx}(5)& =0;\\ \frac{dy}{dx}+1+0& =0.\end{array}}$$

Solving for ⁠dy/dx⁠ gives

$${\displaystyle \frac{dy}{dx}=-1,}$$

the same answer as obtained previously.

Example 2

An example of an implicit function for which implicit differentiation is easier than using explicit differentiation is the function y(x) defined by the equation

$${\displaystyle x^4+2y^2=8.}$$

To differentiate this explicitly with respect to x, one has first to get

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

and then differentiate this function. This creates two derivatives: one for y ≥ 0 and another for y < 0.

It is substantially easier to implicitly differentiate the original equation:

$${\displaystyle 4x^3+4y\frac{dy}{dx}=0,}$$

giving

$${\displaystyle \frac{dy}{dx}=\frac{-4x^3}{4y}=-\frac{x^3}{y}.}$$

Example 3

Often, it is difficult or impossible to solve explicitly for y, and implicit differentiation is the only feasible method of differentiation. An example is the equation

$${\displaystyle y^5-y=x.}$$

It is impossible to express y explicitly in radicals as a function of x, and therefore one cannot find ⁠dy/dx⁠ by explicit differentiation of elementary radical expressions. Using the implicit method, ⁠dy/dx⁠ can be obtained by differentiating the equation to obtain

$${\displaystyle 5y^4\frac{dy}{dx}-\frac{dy}{dx}=\frac{dx}{dx},}$$

where ⁠dx/dx⁠ = 1. Factoring out ⁠dy/dx⁠ shows that

$${\displaystyle \left(5y^4-1\right)\frac{dy}{dx}=1,}$$

which yields the result

$${\displaystyle \frac{dy}{dx}=\frac{1}{5y^4-1},}$$

which is defined for

$${\displaystyle y\ne \pm \frac{1}{\sqrt[4]{5}}\text{and}y\ne \pm \frac{i}{\sqrt[4]{5}}.}$$

Higher order derivatives

The method of implicit differentiation can be applied to find higher-order derivatives such as the second derivative of the implicitly-defined function ${\displaystyle y(x)}$ at the point of its graph ${\displaystyle (x_0,y_0)}$, provided the function ${\displaystyle F}$ is sufficiently smooth at the point.

For example, suppose that the first derivative of ${\displaystyle y(x)}$ has been found, by solving the equation $${\displaystyle dF=F_x(x_0,y_0)dx+F_y(x_0,y_0)dy=0}$$ for ${\displaystyle dy/dx}$ at ${\displaystyle (x_0,y_0)}$.

Applying the second differential to ${\displaystyle F(x,y)=F(x_0,y_0)}$ then gives $${\displaystyle d^{2}F=0}$$ and expanding the left-hand side gives $${\displaystyle d(F_{x}dx+F_{y}dy)=F_{xx}d{x}^2+F_{yx}dxdy+F_{xy}dydx+F_{yy}d{y}^2+F_x{d}^{2}x+F_y{d}^{2}y=0.}$$ where we have written the second partial derivatives of ${\displaystyle F}$ with repeated subscripts. Since it is the ${\displaystyle x}$ derivatives we are after, ${\displaystyle d^{2}x=0}$. Also, substituting the known expressions for ${\displaystyle dy}$ at the point ${\displaystyle (x_0,y_0)}$, into the formula (that is ${\displaystyle dy=\frac{dy}{dx}dx}$ where the derivative on the right-hand side is known at the point ${\displaystyle (x_0,y_0)}$: ${\displaystyle dy/d{x}_{(x_0,y_0)}=-F_x(x_0,y_0)/F_y(x_0,y_0)}$), we can then arrange terms and solve for the second derivative ${\displaystyle d^{2}y/d{x}^2}$. That is, from $${\displaystyle F_{xx}d{x}^2+(F_{xy}+F_{yx})\frac{dy}{dx}d{x}^2+F_{yy}d{y}^2+F_y{d}^{2}y=0}$$ we obtain $${\displaystyle \frac{d^{2}y}{d{x}^2}=-\frac{F_{xx}+(F_{xy}+F_{yx})\frac{dy}{dx}+F_{yy}{\left(\frac{dy}{dx}\right)}^2}{F_y}.}$$ This is understood as at the point ${\displaystyle (x_0,y_0)}$, and requires only that the function ${\displaystyle F}$ be twice-differentiable at the point, in addition to the hypothesis ${\displaystyle F_y(x_0,y_0)\ne 0}$. Second-differentiability at the point is true if the first and second partials of ${\displaystyle F}$ exist and are continuous in a disc around the point.

In that case, Clairaut's theorem implies that the mixed partials are equal ${\displaystyle F_{xy}(x_0,y_0)=F_{yx}(x_0,y_0)}$, and these terms can then be combined to give $${\displaystyle \frac{d^{2}y}{d{x}^2}=-\frac{F_{xx}-2F_{xy}\frac{F_x}{F_y}+F_{yy}{\left(\frac{F_x}{F_y}\right)}^2}{F_y}.}$$

Higher order derivatives (under the requisite smoothness hypotheses) are handled similarly.