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Implicit Differentiation Made Simple

Standard differentiation requires equations to be solved for y (explicit functions like $y=x^2$). Implicit differentiation allows you to find the slope $dy/dx$ even when x and y are mixed together in a messy relationship.

Core Theorem
Chain Rule for Implicit Terms: When differentiating a term with yy with respect to xx, you must multiply by dydx\frac{dy}{dx} (or y′y'). For example, ddx(y3)=3y2⋅y′\frac{d}{dx}(y^3) = 3y^2 \cdot y'.

Practice Exercises

Example 01Easy
Find the tangent line to the circle x2+y2=25x^2 + y^2 = 25 at the point (3,−4)(3, -4).
NEED A HINT?
Differentiate both sides with respect to xx. Remember that the derivative of a constant (25) is 0.
SHOW DETAILED EXPLANATION

Step 1: Differentiate Both Sides

ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25). This gives 2x+2y⋅y′=02x + 2y \cdot y' = 0.

Step 2: Isolate y'

Move 2x2x to the right: 2yy′=−2x2yy' = -2x. Divide by 2y2y: y′=−xyy' = -\frac{x}{y}.

Step 3: Plug in the Point

At (3,−4)(3, -4), the slope is y′=−3−4=34y' = -\frac{3}{-4} = \frac{3}{4}. The tangent line is y−(−4)=34(x−3)y - (-4) = \frac{3}{4}(x-3).
Example 02Hard
Find y′y' for the Folium of Descartes: x3+y3=6xyx^3 + y^3 = 6xy.
NEED A HINT?
The term 6xy6xy requires the **Product Rule** because both xx and yy are variables. Treat it as (6x)⋅(y)(6x) \cdot (y).
SHOW DETAILED EXPLANATION

Step 1: Differentiate Left Side

ddx(x3+y3)=3x2+3y2y′\frac{d}{dx}(x^3 + y^3) = 3x^2 + 3y^2 y'. (Don't forget the chain rule on y3y^3!).

Step 2: Differentiate Right Side (Product Rule)

ddx(6xy)=6(1⋅y+x⋅y′)=6y+6xy′\frac{d}{dx}(6xy) = 6(1 \cdot y + x \cdot y') = 6y + 6xy'.

Step 3: Group y' Terms

3x2+3y2y′=6y+6xy′3x^2 + 3y^2 y' = 6y + 6xy'. Move y′y' terms to one side: 3y2y′−6xy′=6y−3x23y^2 y' - 6xy' = 6y - 3x^2.

Step 4: Solve for y'

Factor out y′y': y′(3y2−6x)=6y−3x2y'(3y^2 - 6x) = 6y - 3x^2. Result: y′=6y−3x23y2−6x=2y−x2y2−2xy' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.
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