Implicit Differentiation Made Simple

Not every equation can be written neatly as \( y = f(x) \). Curves like circles, ellipses and the folium of Descartes mix \( x \) and \( y \) together, and that is exactly where implicit differentiation comes in. Once you understand one simple idea, the chain rule applied to \( y \), the whole method becomes routine.

Explicit vs Implicit Functions

An explicit function gives \( y \) directly in terms of \( x \), for example \( y = 3x^2 + 5 \). An implicit relation connects \( x \) and \( y \) in a single equation, for example:

$$ x^2 + y^2 = 25 \qquad\text{or}\qquad x^3 + y^3 = 6xy $$

You could solve the circle for \( y = \pm\sqrt{25 - x^2} \), but that splits it into two pieces and gets messy. For the second equation, solving for \( y \) is practically impossible. Implicit differentiation lets you find \( \frac{dy}{dx} \) without ever isolating \( y \).

The Key Idea: Treat \( y \) as a Function of \( x \)

When you differentiate with respect to \( x \), remember that \( y \) secretly depends on \( x \). So by the chain rule:

$$ \frac{d}{dx}\big(y^n\big) = n y^{n-1} \frac{dy}{dx} $$

In words: differentiate \( y \) terms as normal, then multiply by \( \frac{dy}{dx} \). A few quick results:

• \( \frac{d}{dx}(y^2) = 2y\frac{dy}{dx} \)
• \( \frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx} \)
• \( \frac{d}{dx}(e^y) = e^y \frac{dy}{dx} \)
• \( \frac{d}{dx}(xy) = y + x\frac{dy}{dx} \) (product rule)

The Four-Step Method

1. Differentiate both sides of the equation with respect to \( x \).
2. Attach \( \frac{dy}{dx} \) every time you differentiate a term containing \( y \).
3. Collect all \( \frac{dy}{dx} \) terms on one side and everything else on the other.
4. Factor out \( \frac{dy}{dx} \) and divide.

We will write \( y' \) for \( \frac{dy}{dx} \) to keep things tidy.

Worked Example 1: The Circle

Find \( \frac{dy}{dx} \) for \( x^2 + y^2 = 25 \), and the equation of the tangent at \( (3, 4) \).

Step 1: Differentiate both sides.

$$ 2x + 2y\,y' = 0 $$

Step 2: Solve for \( y' \).

$$ 2y\,y' = -2x \quad\Rightarrow\quad y' = -\frac{x}{y} $$

Step 3: Evaluate at \( (3, 4) \). First check the point lies on the curve: \( 9 + 16 = 25 \). Then

$$ y' = -\frac{3}{4} $$

Step 4: Write the tangent line.

$$ y - 4 = -\frac{3}{4}(x - 3) \quad\Rightarrow\quad 4y - 16 = -3x + 9 \quad\Rightarrow\quad 3x + 4y = 25 $$

This makes geometric sense: the radius to \( (3, 4) \) has slope \( \frac{4}{3} \), and the tangent is perpendicular to it.

Worked Example 2: The Folium of Descartes

Find \( \frac{dy}{dx} \) for \( x^3 + y^3 = 6xy \) and evaluate it at \( (3, 3) \).

Step 1: Check the point. \( 27 + 27 = 54 \) and \( 6(3)(3) = 54 \). It lies on the curve.

Step 2: Differentiate both sides. The right side needs the product rule.

$$ 3x^2 + 3y^2\,y' = 6y + 6x\,y' $$

Step 3: Collect the \( y' \) terms.

$$ 3y^2\,y' - 6x\,y' = 6y - 3x^2 $$

Step 4: Factor and divide.

$$ y'(3y^2 - 6x) = 6y - 3x^2 \quad\Rightarrow\quad y' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x} $$

Step 5: Substitute \( (3, 3) \).

$$ y' = \frac{2(3) - 9}{9 - 2(3)} = \frac{-3}{3} = -1 $$

So the tangent at \( (3, 3) \) has slope \( -1 \).

Worked Example 3: A Second Derivative

Find \( \frac{d^2y}{dx^2} \) for \( x^2 + y^2 = 25 \).

From Example 1, \( y' = -\frac{x}{y} \). Differentiate again using the quotient rule, remembering \( y \) depends on \( x \):

$$ y'' = -\frac{y \cdot 1 - x\,y'}{y^2} $$

Substitute \( y' = -\frac{x}{y} \):

$$ y'' = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} $$

Since \( x^2 + y^2 = 25 \) on the curve:

$$ y'' = -\frac{25}{y^3} $$

Using the original equation to simplify like this is a classic exam trick, so watch for it.

Worked Example 4: Mixing in Trigonometry

Find \( \frac{dy}{dx} \) if \( \sin y + x^2 y = 4x \).

Differentiate term by term:

$$ \cos y \; y' + \big(2xy + x^2 y'\big) = 4 $$

Collect \( y' \) terms:

$$ y'(\cos y + x^2) = 4 - 2xy \quad\Rightarrow\quad y' = \frac{4 - 2xy}{\cos y + x^2} $$

Where Implicit Differentiation Is Used

• Tangents and normals to circles, ellipses and hyperbolas in FSc and A Level.
• Related rates problems, such as a ladder sliding down a wall.
• Derivatives of inverse functions, for example proving \( \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1 + x^2} \) from \( \tan y = x \).
• Logarithmic differentiation, which is implicit differentiation applied after taking logs.

Common Mistakes

• Forgetting \( y' \) after differentiating a \( y \) term. This is the number one error.
• Missing the product rule on terms like \( xy \) or \( x^2 y \). The derivative of \( xy \) is not \( y' \); it is \( y + xy' \).
• Differentiating only one side of the equation. Both sides must be differentiated.
• Not checking the point lies on the curve before substituting.
• Leaving \( y' \) inside the answer for \( y'' \). Always substitute the first derivative back in.

Key Takeaways

• Implicit differentiation finds \( \frac{dy}{dx} \) without solving for \( y \).
• Every time you differentiate a \( y \) term, multiply by \( \frac{dy}{dx} \) because of the chain rule.
• Gather the \( \frac{dy}{dx} \) terms, factor, and divide to finish.
• For second derivatives, substitute \( y' \) back and use the original equation to simplify.

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