Implicit differentiation without losing dy/dx
Differentiate an equation whose y-values are not isolated, treating y as a function of x every time it appears.
How to recognize the method, run it, and know when it is the wrong choice.
Reviewed July 11, 2026Differentiate both sides with respect to x, attach y′ whenever a y-expression is differentiated, then solve algebraically for y′.
Why y′ appears
Even when the equation does not solve explicitly for y, the curve still makes y depend on x locally. Differentiating y² therefore requires the Chain Rule and produces 2y y′.
The extra factor records how quickly y changes as x changes along the curve.
A practical workflow
Differentiate term by term, marking each y-dependent term. Move every term containing y′ to one side, factor y′, and divide by the remaining coefficient.
Keep x and y in the final derivative unless the problem asks for a slope at a specific point.
When the derivative fails
If the coefficient of y′ becomes zero, the curve may have a vertical tangent or the local function description may fail. The algebra is revealing geometry, not merely causing a denominator problem.
Find dy/dx for x² + y² = 25.
Differentiate both variables; y² needs the Chain Rule.
Collect the derivative term.
Solve for y′.
Common mistakes
- Writing the derivative of y² as 2y without y′.
- Substituting point coordinates before solving for the derivative.
- Assuming one equation always describes y as a single global function of x.
Three takeaways
- Every differentiated y-term gets a chain factor y′.
- Solve for y′ after differentiating the whole equation.
- A zero denominator often signals a vertical tangent.