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Session Length
~11 min
Adaptive Checks
10 questions
Transfer Probes
4
Implicit differentiation extends the chain rule to equations not explicitly solved for y, enabling differentiation of curves like circles and ellipses. Inverse function differentiation provides systematic formulas for derivatives of inverse trigonometric, exponential, and logarithmic functions. Together with the chain rule for composite functions and higher-order derivatives, these techniques form the core of AP Calculus AB Unit 3: Differentiation of Composite, Implicit, and Inverse Functions.
The chain rule is the gateway to differentiating composite functions. When equations relate x and y implicitly (like $x^2 + y^2 = 25$), implicit differentiation treats y as a function of x and applies the chain rule to every y-term. Inverse function derivatives extend this further.
Higher-order derivatives reveal concavity, inflection points, and acceleration. Second derivatives found via implicit differentiation often require substituting the first derivative back into the expression.
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The derivative of a composite function f(g(x)) equals f'(g(x)) times g'(x). It is the foundational technique for differentiating compositions, implicit equations, and inverse functions.
Example: d/dx[sin(x^3)] = cos(x^3) * 3x^2. The outer function (sin) differentiates to cos, evaluated at the inner function, then multiplied by the derivative of the inner function (3x^2).
A technique for finding dy/dx when y is not isolated as an explicit function of x. Differentiate both sides of the equation with respect to x, applying the chain rule to y-terms (since y depends on x), then solve for dy/dx.
Example: For x^2 + y^2 = 25: differentiate to get 2x + 2y(dy/dx) = 0, then solve for dy/dx = -x/y.
Formulas for derivatives of arcsin, arccos, arctan, and other inverse trig functions, derived using implicit differentiation and the Pythagorean identity.
Example: d/dx[arctan(x)] = 1/(1 + x^2). d/dx[arcsin(x)] = 1/sqrt(1 - x^2). These are frequently tested on the AP exam.
If f and g are inverse functions, then g'(x) = 1/f'(g(x)). This formula lets you find the derivative of an inverse function at a point without finding the inverse explicitly.
Example: If f(3) = 7 and f'(3) = 4, then the inverse function g satisfies g(7) = 3 and g'(7) = 1/f'(g(7)) = 1/f'(3) = 1/4.
The second derivative f''(x) is the derivative of f'(x) and measures concavity and rate of change of slope. In implicit differentiation, finding the second derivative often requires substituting the first derivative expression back in.
Example: For x^2 + y^2 = 25, dy/dx = -x/y. To find d^2y/dx^2, differentiate -x/y using the quotient rule and substitute dy/dx = -x/y back in: d^2y/dx^2 = -25/y^3.
A technique for differentiating functions with variables in both the base and exponent, or products/quotients with many factors. Take the natural log of both sides, differentiate implicitly, then solve for dy/dx.
Example: For y = x^x: ln(y) = x ln(x). Differentiate: (1/y)(dy/dx) = ln(x) + 1. So dy/dx = x^x(ln(x) + 1).
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