Guides · 7 min read
Implicit Differentiation, Explained
Most functions you differentiate are already solved for y: y equals some formula in x, and you apply the rules directly. But plenty of equations relating x and y are never solved for y at all. A circle, an ellipse, or a curve like x cubed plus y cubed equals 6xy has no single tidy formula for y in terms of x, and some equations are impossible to solve for y using elementary functions at all. Implicit differentiation lets you find dy/dx anyway, straight from the original equation, without ever isolating y.
The technique itself is not new calculus — it is the chain rule, applied consistently to every term that involves y, plus a bit of algebra to isolate dy/dx afterward. Once the bookkeeping habit is automatic, implicit differentiation becomes one of the more mechanical topics in the course.
Why You Can't Always Solve for y First
Consider the equation x squared plus y squared equals 25, the equation of a circle of radius 5. Solving for y gives two separate functions: y equals the positive square root of 25 minus x squared for the top half, and y equals the negative square root for the bottom half. You would need to differentiate each half separately, and the split feels arbitrary compared to the single equation you started with.
Now consider an equation like x cubed plus y cubed equals 6xy, a classic curve called the folium of Descartes. Solving for y here means running the cubic formula and tracking three branches, and other equations, such as y plus e to the y equals x, cannot be solved for y in elementary functions at all. If your only tool were explicit differentiation, those slopes would be out of reach or miserable to reach. Implicit differentiation removes the requirement to solve for y at all — it works directly on the equation as given.
The Core Idea: Treat y as a Function of x
The entire method rests on one assumption: y is some function of x, even if you cannot write down its formula. Under that assumption, any term containing y is really a composite function of x, and differentiating it requires the chain rule. When you differentiate y squared with respect to x, you get 2y times dy/dx, not just 2y, because you are differentiating a function of a function.
This is exactly parallel to differentiating the sine of g(x), which gives the cosine of g(x) times g prime of x. Here g(x) is simply y, and g prime of x is dy/dx. Keeping that dy/dx factor attached to every derivative of a y-term is the single habit that makes implicit differentiation work.
Differentiating any power of y with respect to x carries a dy/dx factor from the chain rule.
The Step-by-Step Method
Differentiate both sides of the equation with respect to x, term by term. Apply the ordinary rules to every term in x alone. For every term containing y, apply the chain rule and attach a dy/dx factor. For any term where x and y are multiplied together, apply the product rule, remembering that the derivative of the y-factor still carries its own dy/dx.
Once every term is differentiated, collect all the terms that contain dy/dx on one side of the equation and move every other term to the other side. Factor dy/dx out of its terms, then divide to solve for it explicitly. The result is a formula for dy/dx that is typically written in terms of both x and y — and that is expected, not a sign of an error.
- Differentiate both sides with respect to x, term by term.
- Attach dy/dx to the derivative of every y-term (chain rule).
- Use the product rule for any xy term, keeping dy/dx on the y-part.
- Group the dy/dx terms, factor, then divide to isolate dy/dx.
A Worked Example: The Circle
Return to x squared plus y squared equals 25. Differentiating term by term gives 2x plus 2y times dy/dx equals 0, since the derivative of the constant 25 is zero and the y squared term picked up the chain-rule factor. Solving for dy/dx means moving 2x to the other side and dividing by 2y, which gives dy/dx equals negative x over y.
Notice the answer depends on both coordinates, which makes sense geometrically: the slope of a circle is different at every point, and knowing only the x-coordinate is not enough — you also need to know which point on the circle, top or bottom, you are standing at. Plug in a specific point, such as (3, 4), and dy/dx becomes negative three over four, the actual slope of the tangent line there.
A Worked Example With a Product Term
Now try an equation where x and y are multiplied together, such as xy plus y squared equals 10. The first term needs the product rule: the derivative of xy is 1 times y plus x times dy/dx, since the derivative of x is 1 and the derivative of y is dy/dx. The second term, y squared, differentiates to 2y times dy/dx as before.
Putting the pieces together gives y plus x dy/dx plus 2y dy/dx equals 0. Move the lone y to the other side, factor dy/dx out of the remaining two terms, and divide: dy/dx equals negative y over the quantity x plus 2y. The product rule is the step most commonly skipped when a y-term is multiplied by an x-term, so it is worth double-checking whenever you see two variables multiplied directly.
The product rule applied to a mixed xy term, with the y-factor still carrying its own dy/dx.
Finding the Second Derivative Implicitly
Second derivatives work the same way, one differentiation further. Differentiate the first implicit-derivative equation again with respect to x, remembering that dy/dx is itself a function of x and needs the product or quotient rule wherever it appears multiplied by something else. This produces an equation containing y, dy/dx, and the second derivative.
Substitute the expression you already found for dy/dx back into that equation, then solve algebraically for the second derivative. The result is messier than a first derivative from explicit calculus, but the process is the same two moves repeated: differentiate implicitly, then substitute and simplify.
Frequently asked questions
Why does differentiating a term with y produce a dy/dx factor?
Because implicit differentiation assumes y is a function of x, so any expression containing y is a composite function. The chain rule requires multiplying by the derivative of the inner function, which here is dy/dx.
Is it a problem that my answer for dy/dx contains both x and y?
No, that is the expected form. Unlike explicit differentiation, implicit derivatives are usually functions of both variables because the curve itself is not a single function of x alone.
When do I need the product rule instead of just the chain rule?
Whenever a term multiplies an x-expression by a y-expression, such as xy or x squared y. Differentiate it as a product, and remember the derivative of the y-factor still carries its own dy/dx.
How is implicit differentiation related to related rates?
Related rates problems differentiate an equation with respect to time instead of x, so every changing quantity picks up its own time-derivative factor, such as dx/dt or dy/dt. It is the same chain-rule bookkeeping, just with time as the independent variable instead of x.
Can I always solve the resulting equation explicitly for dy/dx?
Almost always. After differentiating, the equation is linear in dy/dx, since the chain rule only ever multiplies dy/dx by a factor, never raises it to a power, so collecting and factoring isolates it. The one catch is a point where that factor is zero, which makes the division invalid and signals a vertical tangent, where dy/dx does not exist.