Calculus-Master
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Implicit Differentiation

A technique used when $y$ cannot be easily isolated as a function of $x$. It treats $y$ as a 'black box' containing an $x$-expression.

Core Theorem
Differentiate both sides with respect to xx, applying the Chain Rule to any term containing yy: ddx[f(y)]=f′(y)⋅dydx\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}.
Step-by-Step SOP
  1. 1

    Differentiate

    Take the derivative of both sides of the equation with respect to xx.
  2. 2

    Chain Rule for y

    Apply the Chain Rule to all yy terms: ddx(yn)=nyn−1⋅y′\frac{d}{dx}(y^n) = ny^{n-1} \cdot y'.
  3. 3

    Group Terms

    Move all terms containing dydx\frac{dy}{dx} to one side and everything else to the other.
  4. 4

    Factor and Solve

    Factor out dydx\frac{dy}{dx} and divide to isolate it.

Practice Exercises

Example 01Easy
Find the slope of the tangent line to x2+y2=25x^2 + y^2 = 25 at the point (3, 4).
NEED A HINT?
Remember that the derivative of a constant (25) is 0, and yy requires the Chain Rule.
SHOW DETAILED EXPLANATION

Step 1: Differentiate both sides

ddx(x2)+ddx(y2)=ddx(25)→2x+2y⋅dydx=0\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25) \rightarrow 2x + 2y \cdot \frac{dy}{dx} = 0.

Step 2: Isolate dy/dx

2ydydx=−2x⇒dydx=−xy2y \frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = -\frac{x}{y}.

Step 3: Evaluate at the point

Plug in (3,4)(3, 4): dydx=−34\frac{dy}{dx} = -\frac{3}{4}.
Example 02Hard
Find dy/dxdy/dx for the curve x3+y3=6xyx^3 + y^3 = 6xy.
NEED A HINT?
Treat 6xy6xy as a product of two functions: 6x6x and yy. Use the Product Rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.
SHOW DETAILED EXPLANATION

Step 1: Differentiate both sides

ddx(x3)+ddx(y3)=ddx(6xy)→3x2+3y2y′=6y+6xy′\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy) \rightarrow 3x^2 + 3y^2 y' = 6y + 6x y'.

Step 2: Group y' terms

Move all terms with y′y' to the left: 3y2y′−6xy′=6y−3x23y^2 y' - 6x y' = 6y - 3x^2.

Step 3: Factor and Solve

y′(3y2−6x)=6y−3x2⇒y′=6y−3x23y2−6xy'(3y^2 - 6x) = 6y - 3x^2 \Rightarrow y' = \frac{6y - 3x^2}{3y^2 - 6x}.

Step 4: Simplify

Divide everything by 3: dydx=2y−x2y2−2x\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}.
Common Pitfalls
  • ⚠
    The Constant TrapForgetting to differentiate the constant on the right side. (e.g., x2+y2=25x^2 + y^2 = 25 becoming ⋯=25\dots = 25 instead of 00).
  • ⚠
    The Product Rule OversightTerms like xyxy require the Product Rule: ddx(xy)=(1)(y)+(x)(y′)\frac{d}{dx}(xy) = (1)(y) + (x)(y'). Don't just write 1⋅y′1 \cdot y'!
  • ⚠
    Missing the ChainTreating yy as a constant. Remember: yy is a function of xx, so it never disappears without leaving a y′y' behind.
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