Exploring Behaviors of Implicit Relations - AP Calculus AB

For the relation $x^2y^2+x=1$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2y}$ This problem requires implicit differentiation of x²y² + x = 1. For x²y², we apply the product rule treating it as (x²)(y²): 2x·y² + x²·2y(dy/dx), and differentiating x gives 1. The equation becomes 2xy² + 2x²y(dy/dx) + 1 = 0. Solving for dy/dx: 2x²y(dy/dx) = -2xy² - 1, which gives dy/dx = -(2xy² + 1)/(2x²y). Choice B incorrectly omits the 1 in the numerator that comes from differentiating the x term on the left side. When performing implicit differentiation, remember to differentiate every term in the equation, including standalone x terms that don't contain y.
If $\sin(xy)=x+y$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{1-\cos(xy)\,y}{\cos(xy)\,x-1}$ This problem involves implicit differentiation of sin(xy) = x + y, where we must apply the chain rule to a composite function. Differentiating sin(xy) gives cos(xy)·d/dx(xy) = cos(xy)·(y + x(dy/dx)), while the right side gives 1 + dy/dx. The equation becomes cos(xy)·y + cos(xy)·x(dy/dx) = 1 + dy/dx, which rearranges to cos(xy)·x(dy/dx) - dy/dx = 1 - cos(xy)·y. Factoring: dy/dx[cos(xy)·x - 1] = 1 - cos(xy)·y, so dy/dx = (1 - cos(xy)·y)/(cos(xy)·x - 1). Choice C has the wrong sign on the 1 in the numerator, a common error when moving terms across the equation. Remember that when a term moves from left to right (or vice versa), its sign changes.
For $x^2\tan y + y = 3$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $-\dfrac{2x\tan y}{x^2\sec^2 y+1}$ This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating $x^2$ tan y + y = 3, $x^2$ tan y gives $x^2$ $\sec^2$ y dy/dx + 2 x tan y by product and chain, + dy/dx. Dy/dx terms from chain on tan y and standalone y. Group: 2 x tan y + $x^2$ $\sec^2$ y dy/dx + dy/dx = 0, ( $x^2$ $sec^2$ y + 1) dy/dx = - 2 x tan y, dy/dx = - 2 x tan y / ( $x^2$ $sec^2$ y + 1). A tempting distractor like choice C fails by omitting the +1 in the denominator, forgetting the dy/dx from y. Identify implicit differentiation when trig functions of y are multiplied by x powers.
For points $(x,y)$ satisfying $x^2y+y^3=7$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+3y^2}$ This problem requires implicit differentiation to find dy/dx from the equation x²y + y³ = 7. When we differentiate both sides with respect to x, the left side requires the product rule on x²y (giving 2xy + x²(dy/dx)) and the chain rule on y³ (giving 3y²(dy/dx)), while the right side gives 0. Collecting all terms with dy/dx on one side gives us x²(dy/dx) + 3y²(dy/dx) = -2xy, which factors as (x² + 3y²)(dy/dx) = -2xy. A common error is forgetting the product rule on x²y and only getting x²(dy/dx), which would lead to the incorrect answer B. The key strategy is to remember that every y term generates a dy/dx factor when differentiated, and to carefully apply the product rule to mixed terms like x²y.
On the level curve $\ln(x+y)+xy=3$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=-\dfrac{\frac{1}{x+y}+y}{\frac{1}{x+y}+x}$ This problem requires implicit differentiation to find dy/dx for the level curve ln(x + y) + xy = 3. Differentiating gives [1/(x + y)] (1 + dy/dx) + (x dy/dx + y) = 0, using chain rule for ln and product rule for xy. The dy/dx terms are [1/(x + y)] dy/dx + x dy/dx, grouped, while - [1/(x + y)] - y is the other side. Solving for dy/dx gives - ([1/(x + y)] + y) / ([1/(x + y)] + x), as shown. Choice D is a tempting distractor if someone omits the x in the denominator, forgetting the dy/dx from xy. To spot implicit differentiation opportunities, identify logarithmic or product terms involving both variables.
If $\sin(xy)+y=x^2$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}$ This problem involves implicit differentiation of sin(xy) + y = x². When differentiating sin(xy), we use the chain rule to get cos(xy) times the derivative of xy, which by the product rule is y + x(dy/dx), giving us cos(xy)[y + x(dy/dx)]. The left side becomes y·cos(xy) + x·cos(xy)(dy/dx) + dy/dx, and the right side gives 2x. Rearranging to isolate dy/dx: x·cos(xy)(dy/dx) + dy/dx = 2x - y·cos(xy), which factors as dy/dx[x·cos(xy) + 1] = 2x - y·cos(xy). Therefore, dy/dx = (2x - y·cos(xy))/(x·cos(xy) + 1). Choice C incorrectly treats y as constant when differentiating sin(xy), missing the y term in the numerator. Remember that when differentiating composite functions like sin(xy), the chain rule requires differentiating the entire inner function xy.
Given $y^2\sin x + x\cos y=2$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x-x\sin y}$ This problem involves implicit differentiation of y²sin x + x cos y = 2. For y²sin x, we use the product rule: y²·cos x + sin x·2y·dy/dx. For x cos y, we get x·(-sin y)·dy/dx + cos y·1. Setting up: y²cos x + 2y sin x·dy/dx - x sin y·dy/dx + cos y = 0. Collecting dy/dx terms: dy/dx(2y sin x - x sin y) = -y²cos x - cos y. Therefore, dy/dx = (-y²cos x - cos y)/(2y sin x - x sin y). Choice B has a plus sign in the denominator (2y sin x + x sin y), which is incorrect—the negative sign comes from differentiating cos y. Remember: when differentiating products involving both x and y, apply the product rule carefully and track the signs from derivatives of trigonometric functions.
If $x^2+ y^2 +xy=10$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{-2x-y}{2y+x}$ This problem uses implicit differentiation on x² + y² + xy = 10. Differentiating term by term: 2x + 2y·dy/dx + (x·dy/dx + y·1) = 0, where the product rule is applied to xy. Collecting terms: 2x + 2y·dy/dx + x·dy/dx + y = 0. Factoring out dy/dx: dy/dx(2y + x) = -2x - y. Therefore, dy/dx = (-2x - y)/(2y + x). Choice C incorrectly gives -2x/(2y + x), which would result from forgetting that the product rule on xy produces both x·dy/dx and y terms. The recognition pattern: whenever you see mixed terms like xy in an implicit equation, expect both variables to contribute to the numerator of dy/dx.
Given $x^2y+\cos y=4$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{-2xy}{x^2-\sin y}$ This problem uses implicit differentiation on x²y + cos y = 4. Differentiating the left side: the product rule on x²y gives x²·dy/dx + 2x·y, and the chain rule on cos y gives -sin y·dy/dx. The right side differentiates to 0. Combining: x²·dy/dx + 2xy - sin y·dy/dx = 0. Factoring out dy/dx: dy/dx(x² - sin y) = -2xy. Therefore, dy/dx = -2xy/(x² - sin y). Choice C incorrectly drops the y from the numerator, giving -2x/(x² - sin y), which would result from forgetting the product rule on x²y. The strategy for avoiding this error: when you see a product like x²y, immediately recognize that both factors may vary, so the product rule is essential.
If $x^3+y^3=6xy$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{6y-3x^2}{3y^2-6x}$ This problem uses implicit differentiation on the equation x³ + y³ = 6xy. Differentiating both sides with respect to x gives 3x² + 3y²·dy/dx = 6x·dy/dx + 6y, where the product rule is applied to the right side. Rearranging to collect dy/dx terms yields 3y²·dy/dx - 6x·dy/dx = 6y - 3x², which factors as dy/dx(3y² - 6x) = 6y - 3x². Therefore, dy/dx = (6y - 3x²)/(3y² - 6x). Choice E incorrectly has a plus sign in the denominator, which would result from a sign error when moving terms. The strategy is to carefully track signs when collecting dy/dx terms to one side of the equation.
If $x^2y^3+y=2$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $-\dfrac{2xy^3}{3x^2y^2+1}$ This problem involves implicit differentiation of x²y³ + y = 2. Using the product rule on x²y³ gives 2x·y³ + x²·3y²·dy/dx, so the full differentiation yields 2xy³ + 3x²y²·dy/dx + dy/dx = 0. Factoring out dy/dx gives dy/dx(3x²y² + 1) = -2xy³, therefore dy/dx = -2xy³/(3x²y² + 1). Choice B incorrectly omits the +1 in the denominator, forgetting that the term y also produces dy/dx when differentiated. The recognition strategy is to remember that every occurrence of y in the original equation will contribute a dy/dx term after differentiation.
For the curve $x^3+y^3=6xy$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{6y-3x^2}{3y^2-6x}$ This problem involves implicit differentiation of x³ + y³ = 6xy. Differentiating both sides gives 3x² + 3y²·dy/dx = 6x·dy/dx + 6y·1, where the product rule is applied to 6xy. Rearranging to collect dy/dx terms: 3y²·dy/dx - 6x·dy/dx = 6y - 3x². Factoring out dy/dx yields dy/dx(3y² - 6x) = 6y - 3x². Therefore, dy/dx = (6y - 3x²)/(3y² - 6x). Choice E has the same numerator but denominator 6x - 3y², which reverses the signs—a common algebraic error when moving terms. The key insight: when collecting dy/dx terms, be systematic about which terms contain dy/dx and which don't, keeping careful track of signs.
For the curve $x+ y\ln x=6$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=-\dfrac{1+y/x}{\ln x}$ This problem asks for dy/dx from x + y ln x = 6, where y is multiplied by ln x. Differentiating: 1 + (ln x)(dy/dx) + y·(1/x) = 0, using the product rule on y ln x and remembering that d/dx(ln x) = 1/x. Solving for dy/dx: (ln x)(dy/dx) = -1 - y/x, so dy/dx = -(1 + y/x)/(ln x). Choice D incorrectly writes the numerator as (1 + y) instead of (1 + y/x), missing that the derivative of ln x contributes a factor of 1/x to the y term. The strategy for logarithmic products is to carefully apply the product rule, remembering that d/dx(ln x) = 1/x, not just 1.
If $e^{x+y}=xy$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{y-e^{x+y}}{e^{x+y}-x}$ This problem requires implicit differentiation of $e^{x+y}$ = xy. Differentiating the left side using the chain rule gives $e^{x+y}$ ·(1 + dy/dx), while the right side uses the product rule to give x·dy/dx + y. Setting these equal: $e^{x+y}$ ·(1 + dy/dx) = x·dy/dx + y. Expanding and collecting dy/dx terms: $e^{x+y}$ + $e^{x+y}$ ·dy/dx = x·dy/dx + y, which rearranges to dy/dx[ $e^{x+y}$ - x] = y - $e^{x+y}$ . Thus dy/dx = (y - $e^{x+y}$ )/( $e^{x+y}$ - x). Choice B incorrectly has y + $e^{x+y}$ in the numerator, likely from a sign error when moving terms. The recognition strategy: carefully track signs when collecting like terms, especially when terms move across the equals sign.
A curve satisfies $e^{y}+xy=3$ . Find $\dfrac{dy}{dx}$ at a general point $(x,y)$ .
Answer: $\dfrac{dy}{dx}=-\dfrac{y}{e^{y}+x}$ This problem requires implicit differentiation of $e^y$ + xy = 3. When differentiating $e^y$ , we apply the chain rule to get $e^y$ (dy/dx), and when differentiating xy, we use the product rule to get y + x(dy/dx). The equation becomes $e^y$ (dy/dx) + y + x(dy/dx) = 0, since the derivative of the constant 3 is zero. Factoring out dy/dx from the terms that contain it: dy/dx( $e^y$ + x) = -y, which gives us dy/dx = -y/( $e^y$ + x). Choice C incorrectly omits the y term that comes from the product rule when differentiating xy, treating it as if only x varies. When using implicit differentiation, always apply the product rule completely to terms like xy, remembering that both x and y can vary.
For points $(x,y)$ satisfying $x^2y+\ln y=4$ , what is $\dfrac{dy}{dx}$ at a general point on the relation?
Answer: $\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+\frac{1}{y}}$ This problem requires implicit differentiation to find dy/dx for the relation x²y + ln y = 4. Differentiating both sides gives 2x y + x² dy/dx + (1/y) dy/dx = 0, using the product rule for x²y and chain rule for ln y. The dy/dx terms are x² dy/dx and (1/y) dy/dx, grouped on one side, while -2x y is on the other. Solving for dy/dx gives -2x y / (x² + 1/y), as the terms factor out. Choice B is a tempting distractor if someone forgets the dy/dx from ln y, omitting the 1/y in the denominator. To recognize when to apply implicit differentiation, check if the equation includes functions of y or products that prevent easy isolation of y.
If $x\cos y+y\sin x=2$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=\dfrac{y\cos x-\cos y}{x\sin y-\sin x}$ This problem requires implicit differentiation of x·cos(y) + y·sin(x) = 2. For x·cos(y), the product rule gives cos(y) + x·(-sin(y))(dy/dx), and for y·sin(x), we get (dy/dx)·sin(x) + y·cos(x). The complete differentiation yields: cos(y) - x·sin(y)(dy/dx) + (dy/dx)·sin(x) + y·cos(x) = 0. Rearranging to isolate dy/dx terms: dy/dx(sin(x) - x·sin(y)) = -cos(y) - y·cos(x), which gives dy/dx = -(cos(y) + y·cos(x))/(sin(x) - x·sin(y)). This simplifies to dy/dx = (y·cos(x) - cos(y))/(x·sin(y) - sin(x)) after factoring out -1. Choice D incorrectly reverses the order of subtraction in the numerator. To verify trigonometric implicit derivatives, ensure each application of the product and chain rules accounts for all variable dependencies.
For the implicit relation $x\sin y+y\cos x=2$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{dy}{dx}=-\dfrac{\sin y-y\sin x}{x\cos y+\cos x}$ This problem requires implicit differentiation of x sin y + y cos x = 2, involving products of x and y with trigonometric functions. Differentiating x sin y requires the product rule: sin y + x cos y(dy/dx), while differentiating y cos x gives cos x(dy/dx) - y sin x. Setting the total derivative to 0: sin y + x cos y(dy/dx) + cos x(dy/dx) - y sin x = 0, which rearranges to (x cos y + cos x)(dy/dx) = -sin y + y sin x. The common mistake in choice D is getting the wrong sign on y sin x, treating it as if it adds to sin y rather than subtracts. The key insight is that when differentiating y cos x, the derivative of cos x is -sin x, introducing a negative sign that makes the final term +y sin x in our rearrangement.
For the implicit curve $e^{x+y}+xy=4$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?
Answer: $-\dfrac{e^{x+y}+y}{e^{x+y}+x}$ This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating $e^{x+y}$ + xy = 4, $e^{x+y}$ gives $e^{x+y}$ (1 + dy/dx) by chain rule, xy gives x dy/dx + y by product rule. Dy/dx terms arise from chain rule on the exponent and product rule on xy. Group: $e^{x+y}$ + $e^{x+y}$ dy/dx + y + x dy/dx = 0, so ( $e^{x+y}$ + x) dy/dx = - ( $e^{x+y}$ + y), dy/dx = - ( $e^{x+y}$ + y)/( $e^{x+y}$ + x). A tempting distractor like choice C fails by omitting the +x in the denominator, possibly forgetting the dy/dx from xy. For transferable strategy, use implicit differentiation when the curve is defined implicitly and you need the slope without solving for y.
Given $\sin(xy)=x+y$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?
Answer: $\dfrac{1-y\cos(xy)}{x\cos(xy)-1}$ This problem involves implicit differentiation of sin(xy) = x + y. Differentiating the left side requires the chain rule: cos(xy)·(x·dy/dx + y·1), giving us cos(xy)·(x·dy/dx + y) = 1 + dy/dx. Expanding yields cos(xy)·x·dy/dx + cos(xy)·y = 1 + dy/dx, and collecting dy/dx terms gives cos(xy)·x·dy/dx - dy/dx = 1 - cos(xy)·y. Factoring out dy/dx yields dy/dx(cos(xy)·x - 1) = 1 - y·cos(xy), so dy/dx = (1 - y·cos(xy))/(x·cos(xy) - 1). Choice B has the wrong sign in the denominator, which would arise from a sign error when moving the dy/dx term. The recognition strategy is to apply the chain rule to composite trigonometric functions and carefully track signs.

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