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Code Scholars Tutoring · AP Calculus AB

Implicit differentiation and tangent equations

Unit 3 · Topics 3.2, 3.6. Original AP Calculus AB notes, worked example, and self-check.

Understand the idea

When a relation defines y\displaystyle y implicitly as a function of x\displaystyle x, differentiating a term involving y\displaystyle y introduces y′\displaystyle y'. Solve algebraically for the first derivative. To obtain a second derivative, differentiate the first-derivative relation with product, quotient, and chain rules, then substitute known coordinates and slope. A point must satisfy the original relation.

ddx(yn)=nyn−1y′\frac d{dx}(y^n)=ny^{n-1}y'

Worked example

For x2+y2=25\displaystyle x^2+y^2=25, find the tangent at (3,4)\displaystyle (3,4).

Differentiating gives 2x+2yy′=0\displaystyle 2x+2yy'=0, so y′=−x/y\displaystyle y'=-x/y. At (3,4)\displaystyle (3,4) the slope is −3/4\displaystyle -3/4. The tangent is y−4=−34(x−3)\displaystyle y-4=-\frac34(x-3).

Try it yourself

Find y′′\displaystyle y'' at (0,5)\displaystyle (0,5) on the same circle.

Show solution and reasoning

Differentiate 2x+2yy′=0\displaystyle 2x+2yy'=0: 2+2(y′)2+2yy′′=0\displaystyle 2+2(y')^2+2yy''=0. At (0,5)\displaystyle (0,5), y′=0\displaystyle y'=0, so y′′=−1/5\displaystyle y''=-1/5.

Unit 3 MCQNext: Inverse functions and inverse trigonometry →