AP Calculus AB and BC glossary

Implicit Second Derivative

Also called: Second implicit derivative

The implicit second derivative is the result of differentiating dy/dx implicitly a second time. The expression that comes out still contains dy/dx, so the first derivative gets substituted back in and simplified until the answer is written only in terms of x and y.

x2+y2=25 ⟹ d2ydx2=−x2+y2y3=−25y3x^2 + y^2 = 25 \ \Longrightarrow\ \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3} = -\frac{25}{y^3}

Begin from the first derivative you have already solved for, then differentiate it with the quotient or product rule, remembering that every yy you differentiate contributes a factor of dy/dxdy/dx. For x2+y2=25x^2 + y^2 = 25, implicit differentiation gives dy/dx=−x/ydy/dx = -x/y.

d2ydx2=ddx(−xy)=−y−x dydxy2=−y+x2yy2=−x2+y2y3\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y - x\,\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}

Watch for a chance to substitute the original equation back in. Here x2+y2x^2+y^2 appears in the numerator and collapses to 2525, leaving −25/y3-25/y^3. That tidy shape is particular to the circle, though. For x3+y3=1x^3+y^3=1 the same procedure ends at −2x/y5-2x/y^5, which is neither constant nor over y3y^3 and is entirely correct, so the look of the answer is no test of the work.

The mistake

Stopping while dy/dxdy/dx is still in the answer when the question asked for the second derivative in terms of xx and yy. At a specific point you may substitute the numeric slope instead, and that earns full credit, but a prompt that says in terms of xx and yy is not satisfied by an expression that still mentions y′y'.

Appears in: Unit 3: Chain Rule, Implicit, and Inverses