Calculus-Master
Start Free Trial

Free Response · Mastering derivatives of multi-variable equations and their rates of change

Implicit Differentiation & Curve Analysis AP Calculus AB/BC FRQ: Explained with Practice & Scoring Tips

You will typically be asked to show that a derivative dy/dx equals a given expression, use that derivative to find tangent lines or special points (horizontal/vertical), and finally apply the Chain Rule to find rates of change with respect to time (t).

On this page

Key Formulashover to see usage
  • Implicit Differentiationddx[f(y)]=f′(y)⋅dydx\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}
    When to useApply the Chain Rule whenever you differentiate a term containing 'y' with respect to 'x'.
  • Horizontal TangentNumerator of dydx=0\text{Numerator of } \frac{dy}{dx} = 0
    When to useSet the top of your derivative fraction to zero to find where the slope is zero (provided the denominator is not zero).
  • Vertical TangentDenominator of dydx=0\text{Denominator of } \frac{dy}{dx} = 0
    When to useSet the bottom of your derivative fraction to zero to find where the slope is undefined (provided the numerator is not zero).
  • Related Rates (with respect to time)ddt[f(x,y)]=…\frac{d}{dt}[f(x, y)] = \dots
    When to useDifferentiate the entire original equation with respect to 't'. Every variable (x and y) will produce a d/dt term (e.g., dx/dt, dy/dt).
Step-by-Step SOP
  1. 1

    1. Showing the Derivative (The 'Show That' Step)

    Differentiate both sides of the equation. Use Product Rule and Chain Rule carefully. Collect all terms with dy/dx on one side, factor out dy/dx, and divide to isolate it. Match your result with the goal given in the prompt.
  2. 2

    2. Finding Points of Interest

    For horizontal/vertical tangents, solve for one variable (usually y) in the derivative's fraction. **Critical:** You must then plug this back into the *original curve equation* to find the corresponding coordinate.
  3. 3

    3. Linear Approximation

    Identify the given point (x1,y1)(x_1, y_1). Calculate the numerical slope mm by plugging these into your dy/dx formula. Use y−y1=m(x−x1)y - y_1 = m(x - x_1) to estimate a nearby value.
  4. 4

    4. Solving the Related Rates Part

    Switch your focus to time (tt). Differentiate the original equation. Substitute all known values (xx, yy, and the given rate like dx/dtdx/dt) to solve for the unknown rate (dy/dtdy/dt).

Practice Questions

123

Question 1

2025 AP Calculus AB Exam — FRQ 6

Medium
Consider the curve GG defined by the equation y3−y2−y+14x2=0y^3 - y^2 - y + \frac{1}{4}x^2 = 0.
APart A
Easy
Show that dydx=−x2(3y2−2y−1)\frac{dy}{dx} = \frac{-x}{2(3y^2 - 2y - 1)}.
Need a Hint?
Differentiate both sides with respect to xx. Remember to use the chain rule (y′y') for every term containing yy.
Show Solution

Implicit Differentiation

Differentiate the equation term by term: ddx(y3−y2−y+14x2)=ddx(0)  ⟹  3y2dydx−2ydydx−dydx+12x=0\frac{d}{dx}(y^3 - y^2 - y + \frac{1}{4}x^2) = \frac{d}{dx}(0) \implies 3y^2\frac{dy}{dx} - 2y\frac{dy}{dx} - \frac{dy}{dx} + \frac{1}{2}x = 0.

Isolate dy/dx

Factor out dydx\frac{dy}{dx}: (3y2−2y−1)dydx=−x2(3y^2 - 2y - 1)\frac{dy}{dx} = -\frac{x}{2}.

Final Verification

Divide by the coefficient to get dydx=−x2(3y2−2y−1)\frac{dy}{dx} = \frac{-x}{2(3y^2 - 2y - 1)}.

AP Scoring Note — 2 Points

P1: Correct implicit differentiation.
P2: Successful verification/isolation leading to the given expression.
BPart B
Medium
There is a point PP on curve GG near (2,−1)(2, -1) with xx-coordinate 1.61.6. Use the line tangent to the curve at (2,−1)(2, -1) to approximate the yy-coordinate of point PP.
Need a Hint?
Find the numerical slope at (2,−1)(2, -1) using the formula from Part A, then write the equation of the tangent line.
Show Solution

Calculate the Slope

Substitute (x,y)=(2,−1)(x, y) = (2, -1) into dydx\frac{dy}{dx}: −22(3(−1)2−2(−1)−1)=−22(3+2−1)=−14\frac{-2}{2(3(-1)^2 - 2(-1) - 1)} = \frac{-2}{2(3+2-1)} = -\frac{1}{4}.

Tangent Line Approximation

The tangent line equation is y−(−1)=−14(x−2)y - (-1) = -\frac{1}{4}(x - 2). Substitute x=1.6x = 1.6: y≈−1−14(1.6−2)y \approx -1 - \frac{1}{4}(1.6 - 2).

Final Result

y≈−1−14(−0.4)=−1+0.1=−0.9y \approx -1 - \frac{1}{4}(-0.4) = -1 + 0.1 = -0.9.

AP Scoring Note — 2 Points

P3: Correct slope of the tangent line (m=−1/4m = -1/4).
P4: Correct tangent line approximation for yy at x=1.6x = 1.6.
CPart C
Medium
For x>0x > 0 and y>0y > 0, there is a point SS on curve GG at which the line tangent to the curve is vertical. Find the yy-coordinate of point SS. Show the work that leads to your answer.
Need a Hint?
A vertical tangent occurs when the derivative is undefined, which means the denominator of dydx\frac{dy}{dx} must be zero.
Show Solution

Set Denominator to Zero

2(3y2−2y−1)=0  ⟹  3y2−2y−1=02(3y^2 - 2y - 1) = 0 \implies 3y^2 - 2y - 1 = 0.

Solve for y

Factor the quadratic: (3y+1)(y−1)=0(3y + 1)(y - 1) = 0. This gives y=−1/3y = -1/3 or y=1y = 1.

Identify Valid Solution

The problem states y>0y > 0, therefore the yy-coordinate of point SS is y=1y = 1.

AP Scoring Note — 2 Points

P5: Setting the denominator equal to 0.
P6: Solving for yy and identifying the correct value based on the given constraint (y>0y > 0).
DPart D
Hard
A particle moves along curve HH defined by 2xy+ln⁡y=82xy + \ln y = 8. At the instant when the particle is at the point (4,1)(4, 1), dxdt=3\frac{dx}{dt} = 3. Find dydt\frac{dy}{dt} at that instant. Show the work that leads to your answer.
Need a Hint?
This is a Related Rates problem. Differentiate the equation with respect to time tt, treating both xx and yy as functions of tt.
Show Solution

Differentiate with respect to t

Use the product rule for 2xy2xy: 2dxdty+2xdydt+1ydydt=02\frac{dx}{dt}y + 2x\frac{dy}{dt} + \frac{1}{y}\frac{dy}{dt} = 0.

Substitute Known Values

Plug in x=4,y=1,dxdt=3x=4, y=1, \frac{dx}{dt}=3: 2(3)(1)+2(4)dydt+11dydt=02(3)(1) + 2(4)\frac{dy}{dt} + \frac{1}{1}\frac{dy}{dt} = 0.

Solve for dy/dt

6+8dydt+dydt=0  ⟹  6+9dydt=0  ⟹  dydt=−69=−236 + 8\frac{dy}{dt} + \frac{dy}{dt} = 0 \implies 6 + 9\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{6}{9} = -\frac{2}{3}.

AP Scoring Note — 3 Points

P7: Attempt at implicit differentiation with respect to tt.
P8: Correct related rates equation.
P9: Correct final value of dy/dtdy/dt.

Question 2

2024 AP Calculus AB Exam — FRQ 5

Medium
Consider the curve defined by the equation x2+3y+2y2=48x^{2}+3y+2y^{2}=48. It can be shown that dydx=−2x3+4y\frac{dy}{dx}=\frac{-2x}{3+4y}.
aPart a
Easy
There is a point on the curve near (2, 4) with x-coordinate 3. Use the line tangent to the curve at (2, 4) to approximate the y-coordinate of this point.
Need a Hint?
First, find the numerical slope at (2, 4) using the given derivative formula, then apply the tangent line equation.
Show Solution

Find the Slope

Calculate the slope at (2,4)(2, 4): dydx∣(x,y)=(2,4)=−2(2)3+4(4)=−419\frac{dy}{dx}|_{(x,y)=(2,4)} = \frac{-2(2)}{3+4(4)} = -\frac{4}{19}.

Tangent Line Approximation

The tangent line equation is y−4=−419(x−2)y - 4 = -\frac{4}{19}(x - 2). For x=3x = 3: y≈4−419(3−2)=7219y \approx 4 - \frac{4}{19}(3 - 2) = \frac{72}{19}.

AP Scoring Note — 2 Points

P1: Correct slope of tangent line.
P2: Correct approximation at x=3x=3.
bPart b
Medium
Is the horizontal line y=1y = 1 tangent to the curve? Give a reason for your answer.
Need a Hint?
A horizontal tangent must have a slope of zero. Check if any point on the line y=1y=1 results in dydx=0\frac{dy}{dx}=0 and lies on the curve.
Show Solution

Identify Horizontal Tangent Condition

Set dydx=0\frac{dy}{dx} = 0: −2x3+4y=0  ⟹  x=0\frac{-2x}{3+4y} = 0 \implies x = 0. If y=1y=1 is a horizontal tangent, the point must be (0,1)(0, 1).

Verify Point on Curve

Check if (0,1)(0, 1) is on the curve: 02+3(1)+2(1)2=5≠480^{2} + 3(1) + 2(1)^{2} = 5 \neq 48. Therefore, the line y=1y=1 is not tangent to the curve.

AP Scoring Note — 2 Points

P1: Considers dydx=0\frac{dy}{dx} = 0 or identifies x=0x=0.
P2: Provides a correct answer with a valid numerical reason.
cPart c
Easy
The curve intersects the positive x-axis at the point (48,0)(\sqrt{48}, 0). Is the line tangent to the curve at this point vertical? Give a reason for your answer.
Need a Hint?
A tangent line is vertical if the denominator of dydx\frac{dy}{dx} is zero and the numerator is non-zero.
Show Solution

Check for Vertical Tangent Condition

Evaluate the denominator of dydx\frac{dy}{dx} at (48,0)(\sqrt{48}, 0): 3+4(0)=33 + 4(0) = 3. Since the denominator 3≠03 \neq 0, the slope is defined.

Conclusion

Because the slope is defined (−2483-\frac{2\sqrt{48}}{3}), the tangent line is not vertical.

AP Scoring Note — 1 Point

P1: Correct answer ('No') with a reason showing the denominator is non-zero.
dPart d
Hard
For time t≥0t \ge 0, a particle is moving along another curve defined by the equation y3+2xy=24y^{3}+2xy=24. At the instant the particle is at the point (4, 2), the y-coordinate of the particle's position is decreasing at a rate of 2 units per second. At that instant, what is the rate of change of the x-coordinate of the particle's position with respect to time?
Need a Hint?
This is a related rates problem. Differentiate the new equation with respect to tt and substitute x=4,y=2,x=4, y=2, and dydt=−2\frac{dy}{dt}=-2.
Show Solution

Differentiate with respect to t

Using the product rule: 3y2dydt+2dxdty+2xdydt=03y^{2}\frac{dy}{dt} + 2\frac{dx}{dt}y + 2x\frac{dy}{dt} = 0.

Substitute Known Values

Substitute x=4,y=2,dydt=−2x=4, y=2, \frac{dy}{dt}=-2: 3(2)2(−2)+2(2)dxdt+2(4)(−2)=03(2)^{2}(-2) + 2(2)\frac{dx}{dt} + 2(4)(-2) = 0.

Solve for dx/dt

−24+4dxdt−16=0  ⟹  4dxdt=40  ⟹  dxdt=10-24 + 4\frac{dx}{dt} - 16 = 0 \implies 4\frac{dx}{dt} = 40 \implies \frac{dx}{dt} = 10 units per second.

AP Scoring Note — 4 Points

P1: Attempts implicit differentiation.
P2: Correct differentiated equation.
P3: Correctly uses dydt=−2\frac{dy}{dt} = -2.
P4: Correct final answer for dxdt\frac{dx}{dt}.

Question 3

2023 AP Calculus AB Exam — FRQ 6

Hard
Consider the curve given by the equation 6xy=2+y36xy = 2 + y^3.
aPart a
Easy
Show that dydx=2yy2−2x\frac{dy}{dx} = \frac{2y}{y^{2}-2x}.
Need a Hint?
Use implicit differentiation on both sides with respect to xx. Remember the product rule for 6xy6xy.
Show Solution

Implicit Differentiation

Differentiate both sides: ddx(6xy)=ddx(2+y3)  ⟹  6y+6xdydx=3y2dydx\frac{d}{dx}(6xy) = \frac{d}{dx}(2+y^3) \implies 6y + 6x\frac{dy}{dx} = 3y^2\frac{dy}{dx}.

Isolate dy/dx

Group dydx\frac{dy}{dx} terms: 6y=(3y2−6x)dydx  ⟹  dydx=6y3y2−6x6y = (3y^2 - 6x)\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{6y}{3y^2-6x}.

Simplify

Divide numerator and denominator by 3: dydx=2yy2−2x\frac{dy}{dx} = \frac{2y}{y^2-2x}.

AP Scoring Note — 2 Points

P1: Correct implicit differentiation.
P2: Successful verification/isolation.
bPart b
Medium
Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
Need a Hint?
A horizontal tangent requires the numerator of dydx\frac{dy}{dx} to be zero.
Show Solution

Set Numerator to Zero

For a horizontal tangent, 2y=0  ⟹  y=02y = 0 \implies y = 0.

Substitute into Original Equation

Plug y=0y=0 into 6xy=2+y36xy = 2 + y^3: 6x(0)=2+03  ⟹  0=26x(0) = 2 + 0^3 \implies 0 = 2, which is impossible.

Conclusion

Since there is no solution for xx when y=0y=0, no such point exists on the curve.

AP Scoring Note — 2 Points

P1: Sets 2y=02y=0 or dydx=0\frac{dy}{dx}=0.
P2: Correct answer with reason (no solution for xx).
cPart c
Hard
Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
Need a Hint?
A vertical tangent occurs when the denominator of dydx\frac{dy}{dx} is zero.
Show Solution

Set Denominator to Zero

For a vertical tangent, y2−2x=0  ⟹  x=y22y^2 - 2x = 0 \implies x = \frac{y^2}{2}.

Substitute into Original Equation

Plug x=y22x = \frac{y^2}{2} into 6xy=2+y36xy = 2 + y^3: 6(y22)y=2+y3  ⟹  3y3=2+y3  ⟹  2y3=2  ⟹  y=16(\frac{y^2}{2})y = 2 + y^3 \implies 3y^3 = 2 + y^3 \implies 2y^3 = 2 \implies y = 1.

Find x-coordinate

Substitute y=1y=1 into x=y22x = \frac{y^2}{2}: x=122=12x = \frac{1^2}{2} = \frac{1}{2}.

Final Answer

The tangent line is vertical at the point (12,1)(\frac{1}{2}, 1).

AP Scoring Note — 3 Points

P1: Sets y2−2x=0y^2-2x=0.
P2: Substitutes x=y22x=\frac{y^2}{2} into original equation.
P3: Correct point (1/2,1)(1/2, 1).
dPart d
Hard
A particle is moving along the curve. At the instant when the particle is at the point (12,−2)(\frac{1}{2}, -2), its horizontal position is increasing at a rate of dxdt=23\frac{dx}{dt} = \frac{2}{3} unit per second. What is the value of dydt\frac{dy}{dt}, the rate of change of the particle's vertical position, at that instant?
Need a Hint?
This is a Related Rates problem. Differentiate the original equation with respect to tt.
Show Solution

Differentiate with respect to t

Apply ddt\frac{d}{dt} to 6xy=2+y36xy = 2 + y^3: 6dxdty+6xdydt=3y2dydt6\frac{dx}{dt}y + 6x\frac{dy}{dt} = 3y^2\frac{dy}{dt}.

Substitute Known Values

Plug in x=12,y=−2,dxdt=23x=\frac{1}{2}, y=-2, \frac{dx}{dt}=\frac{2}{3}: 6(23)(−2)+6(12)dydt=3(−2)2dydt6(\frac{2}{3})(-2) + 6(\frac{1}{2})\frac{dy}{dt} = 3(-2)^2\frac{dy}{dt}.

Solve for dy/dt

−8+3dydt=12dydt  ⟹  −8=9dydt  ⟹  dydt=−89-8 + 3\frac{dy}{dt} = 12\frac{dy}{dt} \implies -8 = 9\frac{dy}{dt} \implies \frac{dy}{dt} = -\frac{8}{9} unit per second.

AP Scoring Note — 2 Points

P1: Correct implicit differentiation with respect to tt.
P2: Correct final answer for dydt\frac{dy}{dt}.

Timed Practice

Ready to challenge yourself?

Take a 20-minute timed mock exam to test your understanding.

Start Timed Test (20m)
Common Pitfalls
  • ⚠
    The Product Rule TrapIn terms like 6xy6xy or x2y2x^2y^2, you MUST use the Product Rule: d/dx(uv)=u′v+uv′d/dx(uv) = u'v + uv'. Forgetting this is the most common way to lose the first 2 points.
  • ⚠
    Constant DifferentiationThe derivative of a constant (e.g., the '48' or '24' on the right side of the equation) is **0**. Do not leave the constant in your differentiated equation.
  • ⚠
    Missing the 'On the Curve' CheckWhen finding a vertical tangent, if you find y=1y=1, you are not done. You must find the xx that goes with y=1y=1 on the original curve. If no such xx exists, there is no tangent there.
  • ⚠
    Sign Errors in d/dtIn related rates, if a coordinate is 'decreasing,' its rate must be entered as a **negative** number (e.g., dy/dt=−2dy/dt = -2).
  • ⚠
    Notation ClarityClearly distinguish between dy/dxdy/dx (slope) and dy/dtdy/dt (velocity). Mixing these up in the related rates section will lead to incorrect substitution.

You Might Also Like