AP Calculus AB Implicit Differentiation Worksheet Solutions (PDF)
This implicit practice solutions for AP Calculus AB covers implicit differentiation (Common Core AP CALC AB 3.2), with a free PDF to print.
This answer key gives a full worked solution for every problem in the implicit differentiation worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.
AP Calculus AB · Common Core State Standards: AP CALC AB 3.2
Implicit practice solutions: exercises 1 to 10
Exercise 1 – Circle
Differentiate: \(2x+2y\,y^{\prime}=0\), so \(y^{\prime}=-\frac{x}{y}\).
At \((3,4)\): \(-\frac{3}{4}\). At \((-3,4)\): \(-\frac{-3}{4}=\frac{3}{4}\).
Final answer: \(\frac{dy}{dx}=-\frac{x}{y};\ -\frac{3}{4}\text{ and }\frac{3}{4}\).
Exercise 2 – Product of x and y
Product rule: \(y+x\,y^{\prime}=0\), so \(y^{\prime}=-\frac{y}{x}\).
Final answer: \(-\frac{y}{x}\).
Exercise 3 – Sum of cubes
\(3x^{2}+3y^{2}y^{\prime}=0\), so \(y^{\prime}=-\frac{x^{2}}{y^{2}}\).
Final answer: \(-\frac{x^{2}}{y^{2}}\).
Exercise 4 – Mixed terms
Differentiate: \(2x+(y+xy^{\prime})+2yy^{\prime}=0\).
Collect: \(y^{\prime}(x+2y)=-(2x+y)\).
Final answer: \(-\frac{2x+y}{x+2y}\).
Exercise 5 – Sine of y
\(\cos y\cdot y^{\prime}=1\), so \(y^{\prime}=\frac{1}{\cos y}=\sec y\).
Final answer: \(\sec y\).
Exercise 6 – Exponential in y
\(e^{y}y^{\prime}=1+y^{\prime}\). Collect: \(y^{\prime}(e^{y}-1)=1\).
Final answer: \(\frac{1}{e^{y}-1}\).
Exercise 7 – Semicubical parabola
\(2yy^{\prime}=3x^{2}\).
Final answer: \(\frac{3x^{2}}{2y}\).
Exercise 8 – Product and a cube
Differentiate: \(2xy+x^{2}y^{\prime}+3y^{2}y^{\prime}=0\).
Collect: \(y^{\prime}(x^{2}+3y^{2})=-2xy\).
Final answer: \(-\frac{2xy}{x^{2}+3y^{2}}\).
Exercise 9 – Slope at a point
Check: \(4+2+1=7\). From the earlier formula, \(y^{\prime}=-\frac{2x+y}{x+2y}\).
At \((2,1)\): \(-\frac{4+1}{2+2}=-\frac{5}{4}\).
Final answer: \(-\frac{5}{4}\).
Exercise 10 – Tangent line to an implicit curve
Check: \(3\cdot 4=12\). Differentiate: \(y^{2}+2xyy^{\prime}=0\), so \(y^{\prime}=-\frac{y}{2x}\).
At \((3,2)\): \(-\frac{2}{6}=-\frac{1}{3}\). Tangent: \(y-2=-\frac{1}{3}(x-3)\).
Final answer: \(y=-\frac{1}{3}x+3\).
Exercises 11 and up
Exercise 11 – Horizontal and vertical tangents
\(18x+8yy^{\prime}=0\), so \(y^{\prime}=-\frac{9x}{4y}\).
Horizontal: numerator zero, \(x=0\), so \(4y^{2}=36\) and \(y=\pm 3\).
Vertical: denominator zero, \(y=0\), so \(9x^{2}=36\) and \(x=\pm 2\).
Final answer: horizontal at \((0,3)\), \((0,-3)\); vertical at \((2,0)\), \((-2,0)\).
Exercise 12 – A shifted circle
Differentiate: \(2x+2yy^{\prime}-4+6y^{\prime}=0\), so \(y^{\prime}=\frac{4-2x}{2y+6}=\frac{2-x}{y+3}\).
Horizontal: \(2-x=0\), so \(x=2\). Substitute: \(4+y^{2}-8+6y=12\), so \(y^{2}+6y-16=0\) and \((y+8)(y-2)=0\).
Final answer: \((2,2)\text{ and }(2,-8)\).
Exercise 13 – Normal line
The tangent slope is \(-\frac{3}{4}\), so the normal slope is the negative reciprocal \(\frac{4}{3}\).
\(y-4=\frac{4}{3}(x-3)\) simplifies to \(y=\frac{4}{3}x\), a line through the center.
Final answer: \(y=\frac{4}{3}x\).
Exercise 14 – Logarithm in the equation
Rewrite: \(\ln x+\ln y=x\). Differentiate: \(\frac{1}{x}+\frac{y^{\prime}}{y}=1\).
Solve: \(\frac{y^{\prime}}{y}=1-\frac{1}{x}\), so \(y^{\prime}=y\left(1-\frac{1}{x}\right)\).
Final answer: \(y\left(1-\frac{1}{x}\right)\).
Exercise 15 – Sine of a product
Chain and product rules: \(\cos(xy)(y+xy^{\prime})=1\).
So \(y+xy^{\prime}=\sec(xy)\) and \(y^{\prime}=\frac{\sec(xy)-y}{x}\).
Final answer: \(\frac{\sec(xy)-y}{x}\).
Exercise 16 – Second derivative
From \(y^{\prime}=-\frac{x}{y}\), differentiate with the quotient rule: \(y^{\prime\prime}=-\frac{y-x\,y^{\prime}}{y^{2}}\).
Substitute \(y^{\prime}=-\frac{x}{y}\): \(y^{\prime\prime}=-\frac{y+\frac{x^{2}}{y}}{y^{2}}=-\frac{y^{2}+x^{2}}{y^{3}}=-\frac{25}{y^{3}}\).
Final answer: \(-\frac{25}{y^{3}}\).
Exercise 17 – Vertical tangents
\(2x-y-xy^{\prime}+2yy^{\prime}=0\), so \(y^{\prime}=\frac{y-2x}{2y-x}\).
Vertical when \(2y-x=0\), that is \(x=2y\). Substitute: \(4y^{2}-2y^{2}+y^{2}=3\), so \(y^{2}=1\).
Final answer: \((2,1)\text{ and }(-2,-1)\).
Exercise 18 – Drone flight path
\(y^{\prime}=-\frac{x}{y}=-\frac{6}{8}=-\frac{3}{4}\).
\(y-8=-\frac{3}{4}(x-6)\), which is \(y=-\frac{3}{4}x+\frac{25}{2}\).
Final answer: \(\text{slope }-\frac{3}{4};\ y=-\frac{3}{4}x+12.5\).
Exercise 19 – Running track
Check: \(4\cdot\frac{81}{25}+9\cdot\frac{64}{25}=\frac{324+576}{25}=36\).
\(8x+18yy^{\prime}=0\), so \(y^{\prime}=-\frac{4x}{9y}=-\frac{4\cdot 9/5}{9\cdot 8/5}=-\frac{1}{2}\).
Final answer: \(-\frac{1}{2}\).
Exercise 20 – Horizontal tangent of a tilted curve
From \(y^{\prime}=-\frac{2x+y}{x+2y}\), the tangent is horizontal when \(2x+y=0\), so \(y=-2x\).
Substitute: \(x^{2}-2x^{2}+4x^{2}=7\), so \(3x^{2}=7\) and \(x=\pm\sqrt{\frac{7}{3}}\).
Final answer: \(\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)\text{ and }\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)\).
Exercise 21 – Derivative of a fractional power
\(3y^{2}y^{\prime}=2x\), so \(y^{\prime}=\frac{2x}{3y^{2}}\).
Substitute \(y=x^{2/3}\): \(y^{\prime}=\frac{2x}{3x^{4/3}}=\frac{2}{3}x^{-1/3}\).
The power rule gives \(\frac{2}{3}x^{2/3-1}=\frac{2}{3}x^{-1/3}\). They agree.
Final answer: \(\frac{2x}{3y^{2}}=\frac{2}{3}x^{-1/3}\).
Key facts behind this implicit practice solutions
- Differentiate every term with respect to x; a y term gets a factor dy/dx.
- Use the product rule for terms such as xy, and the chain rule for sin y or e^y.
- The derivative may contain both x and y; substitute a point on the curve to get a slope.
- Horizontal tangents: numerator zero. Vertical tangents: denominator zero.
How to use this implicit practice solutions
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
Download the implicit practice solutions as a PDF with the button below and keep it with your notes.
Download the implicit practice solutions as a PDF with the button below and keep it with your notes.
Download the implicit practice solutions as a PDF with the button below and keep it with your notes.
Download the implicit practice solutions as a PDF with the button below and keep it with your notes.
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