Implicit Differentiation: Step-by-Step Method with Examples

Implicit differentiation is a calculus technique. You use it to find the derivative \( \frac{dy}{dx} \). This applies when an equation combines \( x \) and \( y \). It does not isolate \( y \) on one side. You take the derivative of both sides with respect to \( x \). Also, apply the chain rule to every term with \( y \). Then, you can isolate \( \frac{dy}{dx} \) using basic algebra.

This method is essential. You use it when you cannot explicitly solve for \( y \). Also, it helps when solving introduces complicated square roots. It can also create multi-part piecewise branches.

Explicit vs. implicit equations

An equation like \( y = x^3 - 4x \) is explicit. Here, \( y \) is already isolated on the left side. So, you can differentiate the right side directly. Use standard power rules for this. However, an equation like \( x^2 + y^2 = 25 \) or \( x^3 + y^3 = 6xy \) is implicit. It defines a geometric curve. In this case, \( x \) and \( y \) are tangled together.

Sometimes, you can convert an implicit equation into explicit formulas. For instance, solve \( x^2 + y^2 = 25 \) for \( y \). This yields two separate functions. They are \( y = \sqrt{25 - x^2} \) for the top half and \( y = -\sqrt{25 - x^2} \) for the bottom half. Differentiating those radical expressions requires the chain rule. You also need to manage multiple positive and negative cases.

In many cases, isolating \( y \) is impossible with standard algebra, such as in \( x^3 + y^3 = 6xy \) or \( \sin(xy) = x \). Implicit differentiation handles all of these curves directly without ever solving for \( y \).

Implicit differentiation is not ideal when \( y \) is already a simple function of \( x \). In those cases, direct explicit differentiation is faster. It also avoids unnecessary algebraic rearrangement.

Why implicit differentiation works

Around most points on a smooth curve, the relation behaves like a function \( y = y(x) \). This is true for a circle or an ellipse. You might not write out its formula. However, if two mathematical expressions remain equal for every point \( (x, y) \), their rates of change must also be equal. This applies to their rates of change with respect to \( x \).

Because \( y \) depends on \( x \), you must treat \( y \) as an inner function. This applies when you differentiate any term containing \( y \). By the chain rule, the derivative of \( [y(x)]^n \) with respect to \( x \) is \( n y^{n-1} \cdot \frac{dy}{dx} \). Conversely, you differentiate terms containing only \( x \) normally. This is because \( \frac{dx}{dx} = 1 \).

The 4-step method for implicit differentiation

1. Differentiate both sides: Apply the derivative operator \( \frac{d}{dx} \) to every term. Do this on both sides of the equation. Differentiate with respect to \( x \).
2. Apply the chain rule to y-terms: Whenever you differentiate a term containing \( y \), multiply by \( \frac{dy}{dx} \) (or \( y' \)). Use the product rule or quotient rule whenever \( x \) and \( y \) are multiplied or divided together.
3. Group terms with dy/dx: Move every term containing \( \frac{dy}{dx} \) to one side of the equation. Then move all non-\( \frac{dy}{dx} \) terms to the opposite side.
4. Factor out dy/dx and solve: Factor \( \frac{dy}{dx} \) from the grouped terms. Then divide by the remaining expression. This isolates \( \frac{dy}{dx} \).

Example 1: finding the slope of a circle

Find \( \frac{dy}{dx} \) for the circle \( x^2 + y^2 = 25 \). Then, evaluate the slope at the point \( (3, 4) \).

Step 1: Differentiate both sides with respect to \( x \):

$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25) $$

Step 2: Apply the power rule to \( x^2 \). Also, use the chain rule for \( y^2 \). Remember, the derivative of a constant is 0.

$$ 2x + 2y \frac{dy}{dx} = 0 $$

Step 3: Subtract \( 2x \) from both sides:

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

Step 4: Divide by \( 2y \):

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

At the point \( (3, 4) \), substitute \( x = 3 \) and \( y = 4 \):

$$ \left. \frac{dy}{dx} \right|_{(3,4)} = -\frac{3}{4} $$

Verification: Differentiating the explicit top half \( y = \sqrt{25 - x^2} \) directly gives \( \frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}} \). Because \( y = \sqrt{25 - x^2} \), this simplifies to \( -\frac{x}{y} \). Implicit differentiation achieves the exact same derivative without radical algebra.

Geometrically, the line segment from the origin to \( (3, 4) \) has a slope of \( \frac{4}{3} \). The tangent line slope \( -\frac{3}{4} \) is its negative reciprocal, confirming that the tangent line is perpendicular to the radius.

Example 2: using the product rule on the Folium of Descartes

Find \( \frac{dy}{dx} \) for the curve \( x^3 + y^3 = 6xy \) at the point \( (3, 3) \).

Step 1: Differentiate both sides with respect to \( x \):

$$ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy) $$

Step 2: Differentiate the left side directly. On the right side, treat \( 6xy \) as \( 6x \cdot y \). Then apply the product rule.

$$ 3x^2 + 3y^2 \frac{dy}{dx} = 6x \left(\frac{dy}{dx}\right) + 6y(1) $$

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

Step 3: Collect all \( y' \) terms on the left and non-\( y' \) terms on the right:

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

Step 4: Factor out \( y' \) and divide:

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

First confirm that \( (3, 3) \) lies on the curve: \( 3^3 + 3^3 = 27 + 27 = 54 \), and \( 6(3)(3) = 54 \). Substituting \( x = 3 \) and \( y = 3 \) yields:

$$ y' = \frac{2(3) - 3^2}{3^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1 $$

Example 3: handling trigonometric and exponential functions

Find \( y' \) for \( \sin(xy) = x \).

Apply the chain rule to the left side. The outer function is sine. Also, the inner function is the product \( xy \).

$$ \cos(xy) \cdot \frac{d}{dx}(xy) = 1 $$

Apply the product rule to the inner function \( xy \):

$$ \cos(xy) \cdot (y + x y') = 1 $$

Distribute \( \cos(xy) \):

$$ y \cos(xy) + x \cos(xy) y' = 1 $$

Isolate the term containing \( y' \):

$$ x \cos(xy) y' = 1 - y \cos(xy) $$

$$ y' = \frac{1 - y \cos(xy)}{x \cos(xy)} $$

Example 4: finding the equation of a tangent line

Find the equation of the tangent line to \( x^2 y + y^3 = 10 \) at the point \( (3, 1) \).

First verify the point lies on the curve: \( 3^2(1) + 1^3 = 9 + 1 = 10 \). Differentiate both sides using the product rule on \( x^2 y \):

$$ 2xy + x^2 y' + 3y^2 y' = 0 $$

Rather than rearranging complicated symbols, plug in \( x = 3 \) and \( y = 1 \) immediately to solve for the numerical slope:

$$ 2(3)(1) + (3)^2 y' + 3(1)^2 y' = 0 $$

$$ 6 + 9y' + 3y' = 0 quadLongrightarrowquad 12y' = -6 quadLongrightarrowquad y' = -\frac{1}{2} $$

Using point-slope form \( y - y_0 = m(x - x_0) \), write the tangent line equation:

$$ y - 1 = -\frac{1}{2}(x - 3) quadLongrightarrowquad y = -\frac{1}{2}x + \frac{5}{2} $$

Example 5: horizontal and vertical tangent lines

Where does the curve \( x^2 + y^2 = 25 \) have horizontal and vertical tangent lines?

The implicit derivative is \( y' = -\frac{x}{y} \).

• Horizontal tangents occur where the slope is zero. This means \( y' = 0 \). For a rational expression, set the numerator to zero. However, ensure the denominator is non-zero.
  Setting \( x = 0 \) in \( x^2 + y^2 = 25 \) gives \( y^2 = 25 \). This means \( y = \pm 5 \).
  Therefore, the horizontal tangent points are \( (0, 5) \) and \( (0, -5) \).
• Vertical tangents occur when the slope is undefined. This happens when the denominator equals zero. Also, the numerator must be non-zero. For example, set \( y = 0 \) in \( x^2 + y^2 = 25 \). This gives \( x^2 = 25 \), so \( x = \pm 5 \). Therefore, the vertical tangent points are \( (5, 0) \) and \( (-5, 0) \).

Finding second derivatives implicitly

To find the second derivative \( y'' = \frac{d^2 y}{dx^2} \) implicitly, differentiate the first derivative \( y' \) with respect to \( x \). Then, substitute the original formula for \( y' \). Finally, simplify the result using the curve’s original equation.

For the circle \( x^2 + y^2 = 25 \), start with \( y' = -\frac{x}{y} \). Apply the quotient rule:

$$ y'' = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y(1) - x(y')}{y^2} = -\frac{y - x y'}{y^2} $$

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

$$ y'' = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{\frac{y^2 + x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3} $$

Because the original equation states \( x^2 + y^2 = 25 \), substitute 25 into the numerator:

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

This result shows concavity directly: on the upper semicircle where \( y > 0 \), \( y'' < 0 \), meaning the curve is concave down. On the lower semicircle where \( y 0 \), meaning the curve is concave up.

Proving inverse function derivatives

Implicit differentiation is the primary tool used to prove derivative formulas for inverse trigonometric and logarithmic functions.

Natural logarithm function: Let \( y = \ln x \). Rewrite this in exponential form as \( e^y = x \). Differentiate both sides implicitly with respect to \( x \):

$$ e^y \cdot y' = 1 quadLongrightarrowquad y' = \frac{1}{e^y} = \frac{1}{x} $$

This proves that the derivative of ln x is \( \frac{1}{x} \).

Arcsine function: Let \( y = \arcsin x \). Rewrite as \( \sin y = x \). Differentiate implicitly:

$$ \cos y \cdot y' = 1 quadLongrightarrowquad y' = \frac{1}{\cos y} $$

You can use the Pythagorean identity. It states that \( \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} \). Then substitute back to get the formula for arcsin x.

$$ \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}} $$

You can derive the derivative formula for arctan x in the exact same manner.

Common mistakes to avoid

• Forgetting dy/dx on y-terms: Students often write \( \frac{d}{dx}(y^3) = 3y^2 \). However, you must use \( 3y^2 \frac{dy}{dx} \). This is the most common student error. You must multiply by \( \frac{dy}{dx} \) for every differentiation step involving \( y \).
• Skipping the product rule on xy terms: Treating \( xy \) as a single variable causes errors. For example, you might write \( \frac{d}{dx}(xy) = y' \). However, \( xy \) is a product of two functions of \( x \). So, its derivative is \( y + x y' \).
• Leaving constant terms unchanged: The derivative of a constant on the right side of an equation is zero. For example, differentiating \( x^2 + y^2 = 25 \) yields \( 2x + 2y y' = 0 \), not \( 25 \).
• Evaluating at a point off the curve: You must confirm the given point \( (x_0, y_0) \) satisfies the original implicit equation. Do this before you find a numerical slope.
• Stopping before isolating dy/dx: If \( \frac{dy}{dx} \) appears on both sides, you must group all terms. Factor \( \frac{dy}{dx} \) out completely. Then isolate it.

Practical applications

• Related rates problems: In related rates, both variables change over time \( t \). You can differentiate implicitly with respect to \( t \). This connects the speeds of two quantities. For example, consider a sliding ladder. Here, \( x^2 + y^2 = L^2 \) leads to \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \).
• Logarithmic differentiation: You can use logarithmic differentiation for variable-exponent expressions. For example, consider \( f(x) = x^x \). Take the natural logarithm of both sides. This transforms the power into a product: \( \ln y = x \ln x \). Then, differentiating implicitly isolates \( y' \). See Logarithmic differentiation and logarithmic differentiation.
• Physics and geometric curves: Implicit equations define orbits and level curves. They also define pressure maps, ellipses, and hyperbolas. You use implicit differentiation to determine their slopes.

Verify an explicit curve branch with the solver

When you solve an implicit equation for an explicit branch, you can confirm your derivative using the calculation tool. Below is the explicit top branch of the circle \( y = \sqrt{25 - x^2} \):

You can check any isolated branch using the online derivative calculator.

Practice problems with solutions

1. Find \( y' \) for \( x^2 + xy + y^2 = 7 \).
2. Next, find the slope of \( x^2 + xy + y^2 = 7 \) at the point \( (1, 2) \).
3. Find \( y' \) for \( y^2 = x^3 \).
4. Also, find \( y' \) for \( xy = 1 \).
5. Find the slope of \( xy + y^2 = 6 \) at the point \( (1, 2) \).
6. Then, find the points where \( x^2 + y^2 = 25 \) has a vertical tangent line.

Solutions and worked steps:

1. Differentiate: \( 2x + (y + xy') + 2yy' = 0 \Rightarrow y'(x + 2y) = -(2x + y) \Rightarrow y' = -\frac{2x + y}{x + 2y} \).

2. Plug \( x = 1, y = 2 \) into \( y' = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5} \).

3. Differentiate: \( 2y y' = 3x^2 \Rightarrow y' = \frac{3x^2}{2y} \).

4. Differentiate: \( y + xy' = 0 \Rightarrow y' = -\frac{y}{x} \). Since \( y = \frac{1}{x} \), this simplifies explicitly to \( -\frac{1}{x^2} \).

5. Differentiate: \( (y + xy') + 2yy' = 0 \Rightarrow y'(x + 2y) = -y \Rightarrow y' = -\frac{y}{x + 2y} \). At \( (1, 2) \), \( y' = -\frac{2}{1 + 4} = -\frac{2}{5} \).

6. Set denominator of \( y' = -\frac{x}{y} \) to zero: \( y = 0 \). Substitute into \( x^2 + y^2 = 25 \) to get \( x = \pm 5 \). Points are \( (5, 0) \) and \( (-5, 0) \).

Frequently asked questions

Why do we multiply by dy/dx when differentiating y terms?

We multiply by \( \frac{dy}{dx} \) because \( y \) is an unknown function of \( x \). The chain rule requires this step. You multiply the derivative of the outer function with respect to \( y \). Also, multiply by the derivative of \( y \) with respect to \( x \).

Can dy/dx contain y in the final answer?

Yes, implicit derivatives usually depend on both \( x \) and \( y \). To compute a numerical slope, plug in both the x-coordinate and y-coordinate of a point on the curve.

When should I use implicit differentiation instead of explicit differentiation?

Use implicit differentiation when you cannot algebraically isolate \( y \). It also helps if isolating \( y \) creates messy square roots. Furthermore, use it if isolating \( y \) splits a curve into separate top and bottom functions.

Do I need a specific point to solve an implicit differentiation problem?

No, finding the general formula for \( \frac{dy}{dx} \) in terms of \( x \) and \( y \) completes the problem. You only need a specific point if you must compute a numerical slope. Also, you need it for a tangent line equation.

Further reading

• Implicit Differentiation (Paul’s Online Math Notes) — worked examples covering algebraic, exponential, and logarithmic implicit curves.
• 3.8 Implicit Differentiation (OpenStax Calculus Volume 1) — textbook explanations, formal problem-solving strategies, and exercise sets.