Implicit Functions 

In mathematics, especially in calculus, functions are often categorized as explicit or implicit. An implicit function is one where the dependent variable is not isolated on one side of the equation. Instead, the function is defined implicitly by an equation involving both the dependent and independent variables.

For example, the equation:

$x^2+y^2=1$

defines a circle and is an implicit function of y in terms of x.

1.0Explicit and Implicit Functions

Explicit Functions

In an explicit function, the dependent variable is written directly in terms of the independent variable.

Example:
y=2 x+3  is explicit because y is given directly in terms of x.

Implicit Functions

In implicit functions, variables are interdependent and not neatly separated.
Example:
x² + xy + y² = 7 is an implicit relation where y is not isolated.

In many cases, it's impossible or very difficult to express y explicitly, which is why implicit functions are crucial in advanced calculus and multivariable analysis.

2.0Solving Implicit Functions

To solve an implicit function, you often need to apply implicit differentiation, especially when you can't isolate y. Here's how you do it:

Example

Given:
x² + y² = 25

Differentiate both sides with respect to x:

$\begin{array}{rl}& \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}(25)\\ & 2x+2y\frac{dy}{dx}=0\end{array}$

Solve for dy/dx:

$2y\frac{dy}{dx}=-2x\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$

This is the derivative of the implicit function, even though y is not written explicitly.

3.0Implicit Functions in Calculus

In calculus, implicit functions allow us to:

• Differentiate functions without solving for the dependent variable
• Work with curves and shapes not defined as y = f(x)
• Use techniques like partial differentiation in multivariable calculus

Use in Coordinate Geometry

For example, circles, ellipses, and hyperbolas are commonly represented by implicit functions because their equations involve both x and y.

4.0Partial Derivatives of Implicit Functions

In multivariable calculus, we often deal with functions like:

F(x, y, z) = 0

To find the partial derivative of z with respect to x (denoted ∂z/∂x), we treat z as a function of x and y, and apply implicit differentiation:

Formula

If
F(x, y, z) = 0

Then:

$\frac{\partial z}{\partial x}=-\frac{\partial F/\partial x}{\partial F/\partial z}$

This is essential in thermodynamics, physics, economics, and optimization problems where variables are related indirectly.

5.0Applications of Implicit Functions

• Physics: Equations of motion and energy relationships
• Engineering: Stress-strain curves and equilibrium conditions
• Economics: Cost functions, utility functions
• Machine Learning: Constraint optimization problems

6.0Solved Examples on Implicit Functions

Example 1: Given the equation: $x^2+y^2=25,find\frac{dy}{dx}$.

Solution:

Differentiate both sides with respect to x:

$\begin{array}{rl}& \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}(25)\\ & 2x+2y\frac{dy}{dx}=0\\ & \frac{dy}{dx}=\frac{-x}{y}\end{array}$

Example 2: Given the equation: $x^2+xy+y^2=7,find\frac{dy}{dx}$.

Solution

Differentiate implicitly:

$\begin{array}{rl}& \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}(xy)+\frac{d}{dx}\left(y^2\right)=0\\ & 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\end{array}$

Group $\frac{dy}{dx}$ terms:

$x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(x+2y)=-2x-y$

Final answer:

$\frac{dy}{dx}=\frac{-2x-y}{x+2y}$

Example 3: Given the equation: $\sin (xy)+x=y.Find\frac{dy}{dx}$.

Solution:

Differentiate both sides:

$\frac{d}{dx}[\sin (xy)]+\frac{d}{dx}(x)=\frac{d}{dx}(y)$

Using chain rule:

$\cos (xy)\cdot \left(y+x\frac{dy}{dx}\right)+1=\frac{dy}{dx}$ 

Simplify and solve for $\frac{dy}{dx}$:

$\cos (xy)\left(y+x\frac{dy}{dx}\right)+1=\frac{dy}{dx}$

Expand and collect $\frac{dy}{dx}$:

$\cos (xy)y+\cos (xy)x\frac{dy}{dx}+1=\frac{dy}{dx}$

Bring all terms to one side:

$\begin{array}{rl}& \cos (xy)x\frac{dy}{dx}-\frac{dy}{dx}=-\cos (xy)y-1\\ & \begin{array}{rl}& \frac{dy}{dx}(\cos (xy)x-1)=-\cos (xy)y-1\\ & \frac{dy}{dx}=\frac{-\cos (xy)y-1}{\cos (xy)x-1}\end{array}\end{array}$

Example 4: Given the equation: $e^{xy}=x+y$. Find $\frac{dy}{dx}$.

Solution:

Differentiate both sides:

$\frac{d}{dx}\left[e^{xy}\right]=\frac{d}{dx}(x+y)$

Apply chain rule:

$e^{xy}\left(y+x\frac{dy}{dx}\right)=1+\frac{dy}{dx}$

Distribute and simplify:

$e^{xy}y+e^{xy}x\frac{dy}{dx}=1+\frac{dy}{dx}$ 

Group $\frac{dy}{dx}$:

$\begin{array}{rl}& e^{xy}x\frac{dy}{dx}-\frac{dy}{dx}=1-e^{xy}y\\ & \frac{dy}{dx}\left(e^{xy}x-1\right)=1-e^{xy}y\\ & \frac{dy}{dx}=\frac{1-e^{xy}y}{e^{xy}x-1}\end{array}$

Example 5: Given the equation: $\ln \left(x^2+y^2\right)=xy$. Find $\frac{dy}{dx}$.

Solution:

Differentiate both sides:

LHS:

$\frac{1}{x^2+y^2}\cdot \left(2x+2y\frac{dy}{dx}\right)$

RHS:

$\frac{dy}{dx}+y$

Now equating both sides:

$\frac{2x+2y\frac{dy}{dx}}{x^2+y^2}=x\frac{dy}{dx}+y$

Multiply both sides by $x^2+y^2$:

$2x+2y\frac{dy}{dx}=\left(x^2+y^2\right)\left(x\frac{dy}{dx}+y\right)$

Expand RHS:

$2x+2y\frac{dy}{dx}=x\left(x^2+y^2\right)\frac{dy}{dx}+y\left(x^2+y^2\right)$

Group $\frac{dy}{dx}$:

$2y\frac{dy}{dx}-x\left(x^2+y^2\right)\frac{dy}{dx}=y\left(x^2+y^2\right)-2x$

Factor and solve:

$\begin{array}{rl}& \frac{dy}{dx}\left(2y-x\left(x^2+y^2\right)\right)=y\left(x^2+y^2\right)-2x\\ & \frac{dy}{dx}=\frac{y\left(x^2+y^2\right)-2x}{2y-x\left(x^2+y^2\right)}\end{array}$

7.0Practice Questions on Implicit Functions

1. $x^3+y^3=6xy$. Find $\frac{dy}{dx}$.
2. $\cos (x+y)=x^2$. Find $\frac{dy}{dx}$.
3. $x^{2}y+y^2=4x$. Find $\frac{dy}{dx}$.
4. $\ln (xy)=x-y$. Find $\frac{dy}{dx}$.
5. $x^2+y^2+z^2=1$ . Find $\frac{\partial z}{\partial x}$ assuming z is implicitly defined.