Dabour Hub EGEgyptian Curriculum 2026/2027
1-1 (b) Implicit differentiation

1-1 (b) Implicit differentiation

Point 2 of 8 — Implicit Differentiation, Parametric Differentiation and Higher DerivativesQuestion bank: 0 questions (0 with pictures)

02 1-1 (b) Implicit differentiation

The method

  • Differentiate both sides of the equation with respect to x.
  • Treat y as a differentiable function of x and apply the chain rule to every term containing y.
  • Collect all terms containing dydx on one side, factorise, and divide.

The chain-rule step

  • ddx(yn)=n yn−1dydx.
  • ddx(sin⁡y)=cos⁡y dydx and ddx(ey)=ey dydx.
  • ddx(xy)=y+x dydx (product rule).
  • If you differentiate with respect to y instead, the factor dxdy appears.
✍️ Worked examples

Example 1: Find dydx if x2+y2=25, then find the tangent and normal at (3,4).

Solution:

  • Differentiate: 2x+2ydydx=0.
  • So dydx=−xy.
  • At (3,4) the slope is −34.
  • Tangent: y−4=−34(x−3), that is 3x+4y=25.
  • Normal slope is the negative reciprocal, 43, so the normal is y−4=43(x−3).

Important remarks

  • The final formula usually contains both x and y.
  • To find the slope at a given x you must first find the matching y from the original equation.
  • A point must lie on the curve before you substitute it; check it in the equation.
  • A horizontal tangent needs the numerator of dydx equal to zero; a vertical tangent needs the denominator equal to zero.

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