Implicit Differentiation

Till now, we have been differentiating functions given in explicit form as \( y = f(x) \). We understand how to differentiate such functions directly. However, what should we do when the function is not explicitly given but instead is defined implicitly?

In an implicit equation, both \( x \) and \( y \) appear together, and they are intertwined in such a way that solving explicitly for \( y \) in terms of \( x \) is either difficult or impossible. Sometimes, we may be able to isolate \( y \) and then differentiate, but in many cases, this is not feasible.

For example, consider the equation

\[ x^5 + y^5 = 5xy. \]

Here, \( y \) is still a function of \( x \), meaning it depends on \( x \), but we cannot explicitly solve for \( y \) in terms of \( x \) easily. Nevertheless, we should still be able to differentiate the equation because \( y \) behaves as a function of \( x \).

The Idea Behind Implicit Differentiation

The key principle is simple: if two expressions are always equal, then their derivatives must also be equal.

Applying this idea to our example equation,

\[ x^5 + y^5 = 5xy. \]

Taking the derivative on both sides,

\[ \frac{d}{dx} (x^5 + y^5) = \frac{d}{dx} (5xy). \]

Since \( y \) is a function of \( x \), we recognize that differentiating \( y^5 \) requires the chain rule, treating \( y \) as an unknown function of \( x \):

\[ \frac{d}{dx} x^5 + \frac{d}{dx} y^5 = \frac{d}{dx} (5xy). \]

Differentiating each term,

\[ 5x^4 + 5y^4 \frac{dy}{dx} = 5 \frac{d}{dx} (x y). \]

Applying the product rule on \( xy \),

\[ 5x^4 + 5y^4 \frac{dy}{dx} = 5 \left( x \frac{dy}{dx} + y \right). \]

Rearranging to isolate \( \frac{dy}{dx} \),

\[ 5y^4 \frac{dy}{dx} - 5x \frac{dy}{dx} = 5y - 5x^4. \]

Factoring out \( \frac{dy}{dx} \),

\[ (5y^4 - 5x) \frac{dy}{dx} = 5y - 5x^4. \]

\[ \frac{dy}{dx} = \frac{5y - 5x^4}{5y^4 - 5x}. \]

Thus, we have successfully differentiated implicitly, without needing to explicitly solve for \( y \).

General Approach to Implicit Differentiation

Differentiate both sides with respect to \( x \), applying the derivative operator \( \frac{d}{dx} \) to the entire equation. Since \( y \) is implicitly dependent on \( x \), every occurrence of \( y \) must be treated as a function of \( x \), even though its explicit form is unknown.
Apply the chain rule correctly when differentiating terms involving \( y \). Specifically, if \( y \) appears as an exponentiated term, trigonometric function, exponential function, or within any composition, the differentiation must account for the inner dependence on \( x \). That is, when differentiating any function of \( y \), say \( g(y) \), we must apply

\[ \frac{d}{dx} g(y) = g'(y) \frac{dy}{dx}. \]

For example, if differentiating \( y^n \), we obtain

\[ \frac{d}{dx} y^n = n y^{n-1} \frac{dy}{dx}. \]
Use the product or quotient rule wherever necessary. If the equation contains products or quotients of expressions of \( x \) and \( y \), differentiation must respect the product or quotient rule.
Isolate \( \frac{dy}{dx} \) algebraically. Once differentiation is complete, all terms involving \( \frac{dy}{dx} \) should be collected on one side of the equation, and any terms independent of \( \frac{dy}{dx} \) should be moved to the other side. The equation is then solved explicitly for \( \frac{dy}{dx} \), ensuring a well-defined derivative expression in terms of \( x \) and \( y \).

Example

Differentiate the equation \( xy^2 + xy = x^5 \) implicitly and find \( \frac{dy}{dx} \).

Solution:

\[ \frac{d}{dx} (xy^2 + xy) = \frac{d}{dx} x^5. \]

\[ \frac{d}{dx} (xy^2) + \frac{d}{dx} (xy) = 5x^4. \]

Applying the product rule to each term,

\[ x \frac{d}{dx} y^2 + y^2 \frac{d}{dx} x + x \frac{d}{dx} y + y \frac{d}{dx} x = 5x^4. \]

Using the chain rule where necessary,

\[ x (2y \frac{dy}{dx}) + y^2 + x \frac{dy}{dx} + y = 5x^4. \]

Rearranging,

\[ 2xy \frac{dy}{dx} + x \frac{dy}{dx} = 5x^4 - y^2 - y. \]

Factoring \( \frac{dy}{dx} \),

\[ (2xy + x) \frac{dy}{dx} = 5x^4 - y^2 - y. \]

\[ \frac{dy}{dx} = \frac{5x^4 - y^2 - y}{2xy + x}. \]

Example

Differentiate the equation \( \sin(xy) = x + y \) implicitly and find \( \frac{dy}{dx} \).

Solution:

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

Applying the chain rule to \( \sin(xy) \),

\[ \cos(xy) \cdot \frac{d}{dx} (xy) = \frac{d}{dx} x + \frac{d}{dx} y. \]

Using the product rule on \( xy \),

\[ \cos(xy) \cdot (x \frac{dy}{dx} + y) = 1 + \frac{dy}{dx}. \]

Rearranging,

\[ \cos(xy) \cdot x \frac{dy}{dx} - \frac{dy}{dx} = 1 - \cos(xy) \cdot y. \]

Factoring \( \frac{dy}{dx} \),

\[ (\cos(xy) x - 1) \frac{dy}{dx} = 1 - \cos(xy) y. \]

\[ \frac{dy}{dx} = \frac{1 - \cos(xy) y}{\cos(xy) x - 1}. \]

\(\blacksquare\)

We can also differentiate the given function in a different way. Starting with the equation:

\[ \sin(xy) = x + y. \]

Taking the inverse sine on both sides,

\[ xy = \sin^{-1}(x + y). \]

Now differentiating both sides,

\[ \frac{d}{dx} (xy) = \frac{d}{dx} \sin^{-1}(x + y). \]

Applying the product rule on the left,

\[ x \frac{dy}{dx} + y = \frac{1}{\sqrt{1 - (x + y)^2}} \cdot (1 + \frac{dy}{dx}). \]

Rearranging,

\[ x \frac{dy}{dx} + y - \frac{1 + \frac{dy}{dx}}{\sqrt{1 - (x + y)^2}} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ x \frac{dy}{dx} - \frac{\frac{dy}{dx}}{\sqrt{1 - (x + y)^2}} = \frac{1}{\sqrt{1 - (x + y)^2}} - y. \]

Factoring \( \frac{dy}{dx} \),

\[ \left( x - \frac{1}{\sqrt{1 - (x + y)^2}} \right) \frac{dy}{dx} = \frac{1}{\sqrt{1 - (x + y)^2}} - y. \]

\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - (x + y)^2}} - y}{x - \frac{1}{\sqrt{1 - (x + y)^2}}}. \]

This expression is completely different from the one obtained earlier using implicit differentiation directly. However, both are mathematically equivalent, meaning they yield the same numerical values for any valid \( x \) and \( y \).

This highlights an important point: the method of differentiation can significantly affect the form of the derivative expression, even though all correct approaches yield equivalent numerical results.