Implicit Differentiation: Step-by-Step Guide
Implicit differentiation finds dy/dx for an equation in which y is not explicitly solved for in terms of x (such as x^2 + y^2 = 25). The technique: differentiate both sides with respect to x, treating y as a function of x and applying the chain rule whenever y appears, then solve algebraically for dy/dx.
Math Racer skill guide · Calculus 1 · AP Calc AB / College
How to solve it, step by step
- Differentiate both sides with respect to x. Apply d/dx to every term on both sides of the equation. Use whatever differentiation rules are required.\[\frac{d}{dx}[\text{LHS}] = \frac{d}{dx}[\text{RHS}]\]
- Apply the chain rule when y appears. Whenever you differentiate a term involving y, multiply by dy/dx. For example, d/dx[y^2] = 2y * dy/dx.\[\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}\]
- Collect all dy/dx terms on one side. Move every term containing dy/dx to the left side and everything else to the right. Factor dy/dx out.\[\frac{dy}{dx}\,[\dots] = [\dots]\]
- Solve for dy/dx. Divide both sides by the factor multiplying dy/dx to express the derivative explicitly.\[\frac{dy}{dx} = \frac{[\text{x and y terms}]}{[\text{x and y terms}]}\]
Worked example
Find dy/dx if x^2 + y^2 = 25.
Step 1: Differentiate both sides with respect to x. d/dx[x^2 + y^2] = d/dx[25].
Step 2: Apply the chain rule: d/dx[x^2] = 2x and d/dx[y^2] = 2y * dy/dx. The equation becomes 2x + 2y * dy/dx = 0.
Step 3: Isolate the dy/dx term: 2y * dy/dx = -2x.
Step 4: Solve for dy/dx: dy/dx = -2x/(2y) = -x/y.
Step 5: Result: dy/dx = -x/y.
Verification: This is the equation of a circle of radius 5. At (3, 4), dy/dx = -3/4, which matches the slope of the tangent line to the circle at that point.
Tips and common mistakes
- Always include dy/dx (or y') whenever differentiating a term that contains y.
- Use the product rule for terms like xy: d/dx[xy] = y + x*dy/dx.
- Use the quotient rule for x/y or y/x — same rules, just with the chain factor on y.
- After differentiating, the result usually contains both x and y; that is normal for implicit form.
- To find a tangent line, plug specific (x_0, y_0) into your dy/dx expression.
- If the equation can be solved for y explicitly, it's faster — but implicit differentiation always works.
- For the second derivative implicitly, differentiate dy/dx again and substitute dy/dx back in to eliminate it.
- Double-check by parameterizing or plugging the resulting (x, y, dy/dx) triple back into the original equation differentiated.