Exploring Behaviors of Implicit Relations - AP Calculus BC

Given $x^2y+\tan(xy)=0$ , find $\dfrac{dy}{dx}$ at a general point $(x,y)$ .
Answer: $\dfrac{-2xy-y\sec^2(xy)}{x^2+x\sec^2(xy)}$ This problem requires implicit differentiation to find dy/dx for the relation x² y + tan(x y) = 0. Differentiating: 2 x y + x² dy/dx + sec²(x y) (y + x dy/dx) = 0. Dy/dx from product and chain rules. Grouping: 2 x y + y sec²(x y) + dy/dx (x² + x sec²(x y)) = 0, dy/dx = - (2 x y + y sec²) / (x² + x sec²) = - y (2 x + sec²) / [x (x + sec²)], but simplified to choice A. Choice B has positive numerator, from sign error. Spot in trig functions of products.
If $\ln(xy)+y=5$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $-\dfrac{1/x}{1+1/y}$ This problem involves implicit differentiation of ln(xy) + y = 5. When differentiating ln(xy) with respect to x, we use the chain rule: d/dx[ln(xy)] = 1/(xy)·d/dx[xy] = 1/(xy)·(x·dy/dx + y·1) = dy/dx/y + 1/x. The full differentiation gives us dy/dx/y + 1/x + dy/dx = 0. Factoring out dy/dx yields dy/dx(1/y + 1) = -1/x, so dy/dx = -1/x ÷ (1/y + 1) = -1/x ÷ (1 + 1/y), which is choice D. Choice C might tempt students who incorrectly simplify the logarithm derivative. The recognition strategy is to carefully apply the chain rule to composite functions and factor out dy/dx terms systematically.
A curve is given by $\sqrt{x}+\sqrt{y}=5$ ; find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Answer: $-\dfrac{\sqrt{y}}{\sqrt{x}}$ This problem requires implicit differentiation of √x + √y = 5 to find dy/dx. Rewriting as $x^{1/2}$ + $y^{1/2}$ = 5 and differentiating gives (1/2) $x^{-1/2}$ + (1/2) $y^{-1/2}$ (dy/dx) = 0. Multiplying through by 2 to clear fractions: $x^{-1/2}$ + $y^{-1/2}$ (dy/dx) = 0, so $y^{-1/2}$ (dy/dx) = - $x^{-1/2}$ . Therefore, dy/dx = - $x^{-1/2}$ / $y^{-1/2}$ = - $y^{1/2}$ / $x^{1/2}$ = -√y/√x. A common error is to invert the fraction incorrectly, getting -√x/√y instead. The key insight is that when dividing by $y^{-1/2}$ , you multiply by $y^{1/2}$ , and the pattern $x^{-1/2}$ / $y^{-1/2}$ = $y^{1/2}$ / $x^{1/2}$ follows from the rule for dividing powers.
For the implicit relation $x\cos y+y\sin x=1$ , find $\dfrac{dy}{dx}$ at a general point $(x,y)$ .
Answer: $-\dfrac{\cos y+y\cos x}{-x\sin y+\sin x}$ This problem uses implicit differentiation on x·cos(y) + y·sin(x) = 1 to find dy/dx. Differentiating requires the product rule on both terms: d/dx[x·cos(y)] = cos(y) + x·(-sin(y))·(dy/dx) and d/dx[y·sin(x)] = (dy/dx)·sin(x) + y·cos(x). Setting the sum equal to 0: cos(y) - x·sin(y)·(dy/dx) + (dy/dx)·sin(x) + y·cos(x) = 0. Collecting dy/dx terms: (sin(x) - x·sin(y))·(dy/dx) = -(cos(y) + y·cos(x)), so dy/dx = -(cos(y) + y·cos(x))/(sin(x) - x·sin(y)). A common error is forgetting the negative sign when cos(y) appears in the derivative of cos(y), leading to sign errors. The recognition strategy is to carefully track signs through trigonometric derivatives and note that the denominator can be written as -(x·sin(y) - sin(x)) = -x·sin(y) + sin(x).
Points $(x,y)$ satisfy $\sin(xy)+y=4$ . What is $\dfrac{dy}{dx}$ at a general point on the curve?
Answer: $\dfrac{-y\cos(xy)}{x\cos(xy)+1}$ This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the relation $\sin(xy) + y = 4$ . The chain rule brings in $\dfrac{dy}{dx}$ when differentiating composite functions like $\sin(xy)$ , accounting for the inner $xy$ changing with $x$ . Specifically, the derivative of $\sin(xy)$ is $\cos(xy)$ times $(y + x \dfrac{dy}{dx})$ , plus $\dfrac{dy}{dx}$ from the standalone $y$ , equaling zero. Terms are grouped by moving non- $\dfrac{dy}{dx}$ parts to one side, yielding $\dfrac{dy}{dx} = \dfrac{-y \cos(xy)}{x \cos(xy) + 1}$ . Choice E fails as a distractor by having a positive numerator, ignoring the negative sign from solving the equation correctly. For transferable strategy, spot implicit relations when transcendental functions entwine $x$ and $y$ , requiring differentiation without explicit solving.
Points $(x,y)$ satisfy $x\cos y+y\sin x=3$ . What is $\dfrac{dy}{dx}$ at a general point?
Answer: $\dfrac{-\cos y-y\cos x}{-x\sin y+\sin x}$ This problem requires implicit differentiation to find dy/dx for the relation x cos y + y sin x = 3. Chain and product rules introduce dy/dx for trigonometric terms with y. Differentiation yields cos y - x sin y dy/dx + sin x dy/dx + y cos x = 0. Terms are grouped by isolating dy/dx factors, giving (-cos y - y cos x)/(-x sin y + sin x). Choice B fails as a distractor with an incorrect sign in the denominator, altering the expression. Recognize this technique for implicit trig relations where x and y are arguments or coefficients in sine and cosine.
A track satisfies $\ln(x+y)+x^2y=0$ ; find $\dfrac{dy}{dx}$ at an arbitrary point $(x,y)$ .
Answer: $\dfrac{-\frac{1}{x+y}-2xy}{\frac{1}{x+y}+x^2}$ This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the given relation. Differentiating $\ln(x+y)$ gives $\frac{1 + \dfrac{dy}{dx}}{x+y}$ , and $x^2 y$ uses product rule for $2xy + x^2 \dfrac{dy}{dx}$ . $\dfrac{dy}{dx}$ terms arise from chain rule on the log and product rule on the polynomial. Grouping yields $[ \frac{1}{x+y} + x^2 ] \dfrac{dy}{dx} = -[ \frac{1}{x+y} + 2xy ]$ , solving for the expression in A. A tempting distractor like choice C forgets the $x^2$ in the denominator from incomplete collection. To recognize when to use implicit differentiation, check if the equation mixes x and y without solving for y explicitly.
If $x^2+ y\,e^x = 10$ , find $\dfrac{dy}{dx}$ at a general point $(x,y)$ .
Answer: $\dfrac{-2x-ye^x}{e^x}$ This problem involves implicit differentiation of x² + y· $e^x$ = 10. The first term simply gives 2x, while the second term requires the product rule: d/dx[y· $e^x$ ] = (dy/dx)· $e^x$ + y· $e^x$ . The complete differentiation yields: 2x + (dy/dx)· $e^x$ + y· $e^x$ = 0. Solving for dy/dx: (dy/dx)· $e^x$ = -2x - y· $e^x$ , which gives dy/dx = (-2x - $y·e^x$ )/ $e^x$ . Choice C shows the common error of leaving dy/dx in the denominator rather than isolating it. The recognition strategy is to identify which terms will produce dy/dx (only those containing y) and systematically collect them on one side.
A path satisfies $\sin(xy)+y=4$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Answer: $\dfrac{-y\cos(xy)}{x\cos(xy)+1}$ This problem requires implicit differentiation to find dy/dx for the implicitly defined relation sin(xy) + y = 4. When differentiating, composite terms like sin(xy) produce cos(xy)(x dy/dx + y) via the chain and product rules, and y yields dy/dx. These dy/dx terms appear because y is a function of x, requiring the chain rule for nested functions. To solve, group dy/dx terms (x cos(xy) dy/dx + dy/dx) and constants (-y cos(xy)), then isolate dy/dx. A tempting distractor like choice E fails because it leaves dy/dx unsolved in the expression. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.
For $\dfrac{1}{x}+\dfrac{1}{y}=1$ , find $\dfrac{dy}{dx}$ at a general point where $x\ne0,y\ne0$ .
Answer: $-\dfrac{y^2}{x^2}$ This problem requires implicit differentiation to find dy/dx for the relation 1/x + 1/y = 1. Differentiating gives -1/x² - (1/y²) dy/dx = 0. The dy/dx term appears from the chain rule on 1/y, treated as $y^{-1}$ . Solving isolates dy/dx = - (1/x²) / (-1/y²) = - y² / x². Choice A omits the negative sign, possibly from forgetting the chain rule's impact on signs. Identify implicit differentiation when reciprocals or fractions involve both variables.
Given $\arctan(y)=x^2y$ , find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Answer: $\dfrac{2xy}{\frac{1}{1+y^2}-x^2}$ This problem requires implicit differentiation to find dy/dx for the implicitly defined relation arctan y = x² y. When differentiating, terms like arctan y produce (1/(1 + y²)) dy/dx, and x² y yield 2 x y + x² dy/dx. These dy/dx terms appear because y is a function of x, applying chain and product rules. To solve, group dy/dx terms ((1/(1 + y²)) dy/dx - x² dy/dx) and constants (-2 x y), then isolate dy/dx. A tempting distractor like choice D fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.
The relation $x^2+\tan y=3y$ defines $y$ implicitly. Find $\dfrac{dy}{dx}$ .
Answer: $\dfrac{-2x}{\sec^2 y-3}$ This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + tan y = 3y. When differentiating, terms like tan y produce sec² y dy/dx via the chain rule, and 3y yields 3 dy/dx. These dy/dx terms appear because y is a function of x, necessitating the chain rule. To solve, group dy/dx terms (sec² y dy/dx - 3 dy/dx) and constants (-2x), then isolate dy/dx. A tempting distractor like choice D fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.
Given $x^2+xy+\sin y=0$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{-2x-y}{x+\cos y}$ This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + x y + sin y = 0. When differentiating, terms like sin y produce cos y dy/dx, and x y yield x dy/dx + y. These dy/dx terms appear because y is a function of x, applying chain and product rules. To solve, group dy/dx terms (x dy/dx + cos y dy/dx) and constants (-2x - y), then isolate dy/dx. A tempting distractor like choice E fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.
If $e^{x+y}+xy=4$ models a constraint between $x$ and $y$ , what is $\dfrac{dy}{dx}$ ?
Answer: $\dfrac{-e^{x+y}-y}{e^{x+y}+x}$ This problem requires implicit differentiation to find dy/dx for the given relation. Differentiating $e^{x+y}$ introduces $e^{x+y}$ (1 + dy/dx) via the chain rule, and xy uses the product rule for y + x dy/dx. The dy/dx terms emerge from chain and product rules on y-dependent parts. Collecting gives $e^{x+y}$ dy/dx + x dy/dx = - $e^{x+y}$ - y, leading to dy/dx = -( $e^{x+y}$ + y)/( $e^{x+y}$ + x). A tempting distractor like choice D forgets the +x in the denominator from improper grouping. To recognize when to use implicit differentiation, check if the equation mixes x and y without solving for y explicitly.
For $x^2e^y+y=4$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{-2xe^y}{x^2e^y+1}$ This problem requires implicit differentiation to find dy/dx for the relation x² $e^y$ + y = 4. Differentiating both sides with respect to x introduces dy/dx via the chain rule for terms involving y. The term x² $e^y$ requires the product rule, yielding 2x $e^y$ + x² $e^y$ dy/dx, while the standalone y differentiates to dy/dx. Collecting like terms groups the dy/dx factors together as (x² $e^y$ + 1) dy/dx = -2x $e^y$ , allowing us to solve for dy/dx = -2x $e^y$ / (x² $e^y$ + 1). A tempting distractor like choice D forgets the dy/dx from the y term, incorrectly omitting the +1 in the denominator. To recognize implicit differentiation opportunities, look for equations defining y implicitly without solving for y explicitly.
On the curve $x^2+xy+y^2=7$ , determine $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Answer: $\dfrac{-2x-y}{x+2y}$ This problem requires implicit differentiation to find dy/dx for the given relation. Differentiating x², xy, and y² produces 2x, (y + x dy/dx), and 2y dy/dx respectively. The dy/dx appears through product rule on xy and chain rule on y². Grouping terms results in x dy/dx + 2y dy/dx = -2x - y, so dy/dx = -(2x + y)/(x + 2y). A tempting distractor like choice D omits the y in the numerator by missing part of the product rule. To recognize when to use implicit differentiation, check if the equation mixes x and y without solving for y explicitly.
Points $(x,y)$ satisfy $y\,e^{x}+x\,e^{y}=10$ ; find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Answer: $-\dfrac{ye^{x}+e^{y}}{e^{x}+xe^{y}}$ This problem requires implicit differentiation of y· $e^x$ + x· $e^y$ = 10 to find dy/dx. Using the product rule on both terms: d/dx[y· $e^x$ ] = (dy/dx)· $e^x$ + y· $e^x$ and d/dx[x· $e^y$ ] = $e^y$ + x· $e^y$ ·(dy/dx). Setting the sum equal to 0: (dy/dx)· $e^x$ + y· $e^x$ + $e^y$ + x· $e^y$ ·(dy/dx) = 0. Collecting dy/dx terms: $e^x$ ·(dy/dx) + x· $e^y$ ·(dy/dx) = -( $y·e^x$ + $e^y$ ), which factors as ( $e^x$ + $x·e^y$ )(dy/dx) = -( $y·e^x$ + $e^y$ ). Therefore, dy/dx = -( $y·e^x$ + $e^y$ )/( $e^x$ + $x·e^y$ ). A common mistake is to include dy/dx in the denominator of the final answer (as in choice B), forgetting that we've already solved for it. The key insight is that once you've isolated dy/dx algebraically, it shouldn't appear anywhere in your final expression.
For $x^2+ y^2=\ln(xy)$ , find $\dfrac{dy}{dx}$ at a general point where $x>0,y>0$ .
Answer: $\dfrac{\frac{1}{x}-2x}{2y-\frac{1}{y}}$ This problem requires implicit differentiation to find dy/dx for the relation x² + y² = ln(x y). Differentiating: 2 x + 2 y dy/dx = (1/(x y)) (y + x dy/dx) = (1/x + (1/y) dy/dx)/y wait, no: d/dx ln(x y) = 1/(x y) * (y + x dy/dx) = [y + x dy/dx] / (x y) = 1/x + (1/y) dy/dx. So 2x + 2y dy/dx = 1/x + (1/y) dy/dx. Grouping: 2x - 1/x = dy/dx (1/y - 2y), wait: 2x - 1/x + dy/dx (2y - 1/y) = 0? From left - right = 0: 2x + 2y dy/dx - 1/x - (1/y) dy/dx = 0. So (2x - 1/x) + dy/dx (2y - 1/y) = 0, dy/dx = - (2x - 1/x) / (2y - 1/y) = (1/x - 2x) / (2y - 1/y), yes A. Choice B has positive 2x - 1/x. Recognize in logs of products.
If $e^{x+y}+xy=4$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{-e^{x+y}-y}{e^{x+y}+x}$ This problem involves implicit differentiation of $e^{x+y}$ + xy = 4. Differentiating the exponential term requires the chain rule: d/dx[ $e^{x+y}$ ] = $e^{x+y}$ ·(1 + dy/dx). The product term xy gives y + x(dy/dx) by the product rule. Setting up the equation: $e^{x+y}$ (1 + dy/dx) + y + x(dy/dx) = 0. Expanding and collecting dy/dx terms: $e^{x+y}$ + $e^{x+y}$ (dy/dx) + y + x(dy/dx) = 0, which gives [ $e^{x+y}$ + x](dy/dx) = - $e^{x+y}$ - y. Choice C incorrectly leaves dy/dx in the denominator, showing a misunderstanding of solving for dy/dx. The strategy is to expand all terms before collecting dy/dx coefficients.
For points $(x,y)$ on the curve $x^2y+3y^2=6$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{-2xy}{x^2+6y}$ This problem requires implicit differentiation to find dy/dx for the given relation. When differentiating x²y, the product rule introduces a dy/dx term alongside the 2xy term. Differentiating 3y² yields 6y dy/dx due to the chain rule. Collecting dy/dx terms gives x² dy/dx + 6y dy/dx = -2xy, leading to dy/dx = -2xy / (x² + 6y) after factoring. A tempting distractor like choice C omits the 6y in the denominator because someone might forget the chain rule on y². To recognize when to use implicit differentiation, check if the equation mixes x and y without solving for y explicitly.

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