AP Calculus AB: Implicit Differentiation (Drill 1)

Question 1. If \( x^2 + y^2 = 25 \), what is \( \dfrac{dy}{dx} \)?

• A) \( \dfrac{x}{y} \)
• B) \( -\dfrac{x}{y} \) ✓
• C) \( \dfrac{y}{x} \)
• D) \( -\dfrac{y}{x} \)

Explanation: Choice B is correct. Differentiating both sides with respect to \( x \): \( 2x + 2y\dfrac{dy}{dx} = 0 \), so \( \dfrac{dy}{dx} = -\dfrac{x}{y} \). Choice A is incorrect because the student dropped the negative sign when solving for \( \dfrac{dy}{dx} \). Choice C is incorrect because the student inverted the fraction, swapping \( x \) and \( y \) in the ratio, and also dropped the negative sign. Choice D is incorrect because the student inverted the fraction, writing \( -\dfrac{y}{x} \) instead of \( -\dfrac{x}{y} \), keeping the negative but reversing which variable is in the numerator.

Question 2. Find \( \dfrac{dy}{dx} \) if \( 3x^2 - 2y^3 = 7 \).

• A) \( \dfrac{x}{y^2} \) ✓
• B) \( \dfrac{x}{y^3} \)
• C) \( -\dfrac{x}{y^2} \)
• D) \( \dfrac{2x}{3y^2} \)

Explanation: Choice A is correct. Differentiating implicitly: \( 6x - 6y^2\dfrac{dy}{dx} = 0 \). Solving: \( \dfrac{dy}{dx} = \dfrac{6x}{6y^2} = \dfrac{x}{y^2} \). Choice B is incorrect because the student applied the power rule to \( y^3 \) but did not subtract 1 from the exponent, treating the derivative of \( y^3 \) as \( 3y^3 \) rather than \( 3y^2 \), and leaving \( y^3 \) in the denominator. Choice C is incorrect because the student introduced a spurious negative sign; solving \( 6x = 6y^2\dfrac{dy}{dx} \) gives a positive result. Choice D is incorrect because the student used the original coefficients 2 and 3 as a ratio \( \dfrac{2}{3} \) rather than differentiating properly and canceling the common factor of 6.

Question 3. The curve \( x^2 + xy + y^2 = 7 \) passes through the point \( (1, 2) \). What is the slope of the tangent line at \( (1, 2) \)?

• A) \( -\dfrac{4}{5} \) ✓
• B) \( \dfrac{4}{5} \)
• C) \( -\dfrac{5}{4} \)
• D) \( -\dfrac{2}{5} \)

Explanation: Choice A is correct. Differentiating implicitly: \( 2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0 \), where the product rule is required on \( xy \). Collecting terms: \( (x + 2y)\dfrac{dy}{dx} = -(2x + y) \), so \( \dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y} \). At \( (1, 2) \): \( \dfrac{dy}{dx} = -\dfrac{2 + 2}{1 + 4} = -\dfrac{4}{5} \). Choice B is incorrect because the student lost the negative sign when solving for \( \dfrac{dy}{dx} \). Choice C is incorrect because the student inverted the fraction, writing \( -\dfrac{x + 2y}{2x + y} \) instead of \( -\dfrac{2x + y}{x + 2y} \), and evaluated at \( (1, 2) \) to get \( -\dfrac{5}{4} \). Choice D is incorrect because the student omitted the \( y \) term when differentiating \( xy \) via the product rule, writing \( x\dfrac{dy}{dx} \) but dropping the \( +y \) contribution, which gives \( \dfrac{dy}{dx} = -\dfrac{2x}{x + 2y} = -\dfrac{2}{5} \) at \( (1, 2) \).

Question 4. Suppose \( F(x, y) = 0 \) implicitly defines \( y \) as a function of \( x \) near a point where \( \dfrac{dy}{dx} \) exists. Which of the following statements is always true?

• A) \( \dfrac{dy}{dx} \) exists at every point on the curve \( F(x, y) = 0 \).
• B) Implicit differentiation gives the same result as solving for \( y \) explicitly and differentiating, wherever both methods are valid. ✓
• C) The curve \( F(x, y) = 0 \) must pass the vertical line test for implicit differentiation to be applied.
• D) \( \dfrac{dy}{dx} \) computed implicitly gives the slope of the normal line to the curve.

Explanation: Choice B is correct. Implicit differentiation is an application of the chain rule to both sides of \( F(x, y) = 0 \), treating \( y \) as a function of \( x \). Wherever \( y \) can be expressed explicitly as a smooth function of \( x \), both methods yield identical results; this is the foundational justification for the technique. Choice A is incorrect because \( \dfrac{dy}{dx} \) fails to exist where the tangent line is vertical, such as at the leftmost and rightmost points of a circle where the denominator of the implicit derivative equals zero. Choice C is incorrect because implicit differentiation only requires \( y \) to be locally a function of \( x \) near the point of interest, not globally, a full circle fails the vertical line test yet implicit differentiation is valid at most of its points. Choice D is incorrect because \( \dfrac{dy}{dx} \) gives the slope of the tangent line; the normal line has slope \( -\dfrac{1}{dy/dx} \).

Question 5. Given \( x^2 + y^2 = 25 \), find \( \dfrac{d^2y}{dx^2} \) in simplified form.

• A) \( -\dfrac{25}{y^3} \) ✓
• B) \( -\dfrac{1}{y} \)
• C) \( -\dfrac{x^2 - y^2}{y^3} \)
• D) \( \dfrac{25}{y^2} \)

Explanation: Choice A is correct. From the first derivative \( \dfrac{dy}{dx} = -\dfrac{x}{y} \), differentiate again using the quotient rule: \( \dfrac{d^2y}{dx^2} = -\dfrac{y(1) - x\dfrac{dy}{dx}}{y^2} = -\dfrac{y - x\left(-\dfrac{x}{y}\right)}{y^2} = -\dfrac{y + \dfrac{x^2}{y}}{y^2} = -\dfrac{x^2 + y^2}{y^3} \). Substituting \( x^2 + y^2 = 25 \) gives \( -\dfrac{25}{y^3} \). Choice B is incorrect because the student differentiated \( -\dfrac{x}{y} \) as \( -\dfrac{1}{y} \), treating \( y \) as a constant and ignoring the quotient rule entirely. Choice C is incorrect because the student made a sign error when substituting \( \dfrac{dy}{dx} = -\dfrac{x}{y} \) back into the quotient rule expression, effectively using \( +\dfrac{x}{y} \) instead, which produces \( -\dfrac{y^2 - x^2}{y^3} = -\dfrac{-(x^2 - y^2)}{y^3} \), giving the wrong sign pattern in the numerator. Choice D is incorrect because the student substituted \( x^2 + y^2 = 25 \) correctly but made two additional errors: dropping the negative sign and writing \( y^2 \) in the denominator instead of \( y^3 \).