Implicit Differentiation
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AP Calculus AB › Implicit Differentiation
For the curve defined by $x^2+xy+y^2=7$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{2x}{x}$
$\dfrac{dy}{dx}=-\dfrac{2x+xy+y^2}{x^2+x+2y}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+y}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{1+2y}$
Explanation
This problem requires implicit differentiation to find the derivative dy/dx for the implicitly defined curve x² + xy + y² = 7. Differentiating both sides with respect to x yields 2x + (x dy/dx + y) + 2y dy/dx = 0, where the product rule is applied to the xy term. The terms involving dy/dx are x dy/dx + 2y dy/dx, which factor to (x + 2y) dy/dx, while the remaining terms are 2x + y. Solving for dy/dx gives dy/dx = -(2x + y)/(x + 2y), grouping the dy/dx terms on one side and constants on the other. A tempting distractor like choice D fails because it incorrectly simplifies the denominator to x + y instead of x + 2y, likely from mishandling the derivative of y². To recognize when to use implicit differentiation, look for equations where y is not isolated as a function of x but both variables are mixed together.
For the curve $x^2y^2+x+y=0$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2xy}$
$\dfrac{dy}{dx}=-\dfrac{2x^2y^2+1}{2x^2y}$
$\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2y}$
$\dfrac{dy}{dx}=\dfrac{2xy^2+1}{2x^2y}$
$\dfrac{dy}{dx}=-\dfrac{2xy^2}{2x^2y+1}$
Explanation
This problem involves implicit differentiation for x² y² + x + y = 0 to find dy/dx. Differentiating gives 2x y² + x² (2 y dy/dx) + 1 + dy/dx = 0, product on x² y². Dy/dx terms: 2 x² y dy/dx + dy/dx, factoring to (2 x² y + 1) dy/dx. Constants: 2x y² + 1, so dy/dx = - (2x y² + 1) / (2 x² y). Choice C tempts by simplifying denominator to 2xy but fails to include the +1 from derivative of y. Recognize in equations with high-degree products like x² y².
For $x^2+xy=\ln y$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x-\frac{1}{y}}$
$\dfrac{dy}{dx}=-\dfrac{2x}{x-\frac{1}{y}}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+\frac{1}{y}}$
$\dfrac{dy}{dx}=\dfrac{2x+y}{x-\frac{1}{y}}$
Explanation
This problem uses implicit differentiation for x² + x y = ln y to find dy/dx. Differentiating gives 2x + y + x dy/dx = (1/y) dy/dx, product on xy, chain on ln y. Dy/dx terms: x dy/dx - (1/y) dy/dx, grouping to (x - 1/y) dy/dx. Constants: 2x + y, so dy/dx = - (2x + y) / (x - 1/y). Choice B tempts with +1/y but fails due to sign error in denominator. Recognize in logarithmic equations with polynomial terms.
For the curve $x^2y+\tan y=5$, what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$?
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=\dfrac{2xy}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=-\dfrac{2x}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2-\sec^2 y}$
Explanation
This problem uses implicit differentiation for x²y + tan y = 5 to find dy/dx. Differentiating yields 2x y + x² dy/dx + sec² y dy/dx = 0, with product rule on x²y and chain on tan y. Dy/dx terms are x² dy/dx + sec² y dy/dx, factoring to (x² + sec² y) dy/dx. The constant term is 2x y, so dy/dx = -2x y / (x² + sec² y). Choice E tempts by omitting the negative sign but fails as differentiation requires moving positive terms to negative. Recognize this in equations mixing polynomials and trig functions of y.
If $\arctan(y)=x^2y$, what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$?
$\dfrac{dy}{dx}=\dfrac{2xy}{\frac{1}{1+y^2}+x^2}$
$\dfrac{dy}{dx}=\dfrac{2x}{\frac{1}{1+y^2}-x^2}$
$\dfrac{dy}{dx}=\dfrac{2xy}{\frac{1}{1+y^2}}$
$\dfrac{dy}{dx}=\dfrac{2xy}{1+y^2-x^2}$
$\dfrac{dy}{dx}=\dfrac{2xy}{\frac{1}{1+y^2}-x^2}$
Explanation
This problem requires implicit differentiation for arctan(y) = x² y to find dy/dx. Differentiating gives [1/(1 + y²)] dy/dx = 2x y + x² dy/dx, chain on left, product on right. Dy/dx terms: [1/(1 + y²)] dy/dx - x² dy/dx, grouping to [1/(1 + y²) - x²] dy/dx. Equals 2x y, so dy/dx = 2x y / [1/(1 + y²) - x²]. Choice C tempts with +x² but fails due to sign error in denominator. Recognize in inverse trig functions equal to polynomials in x and y.
If $y^2\ln x + x\ln y = 3$, what is $\dfrac{dy}{dx}$ at a general point?
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}+\ln y}{2\ln x+\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{2y}{x}+\ln y}{2y\ln x+\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}}{2y\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y}{\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}+\ln y}{2y\ln x+\frac{x}{y}}$
Explanation
This problem involves implicit differentiation for y² ln x + x ln y = 3 to find dy/dx. Differentiating gives 2y dy/dx ln x + y² (1/x) + ln y + x (1/y) dy/dx = 0, applying product and chain rules. Dy/dx terms are 2y ln x dy/dx + (x/y) dy/dx, factoring to (2y ln x + x/y) dy/dx. Constants are y²/x + ln y, so dy/dx = - (y²/x + ln y) / (2y ln x + x/y). Choice B is tempting but fails by omitting y in the denominator's first term, likely a chain rule error. Recognize this in logarithmic products with both x and y.
For the relation $x\ln y+y\ln x=10$, what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=\dfrac{\ln y+\frac{y}{x}}{\frac{x}{y}+\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{y}{x}}{\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{x}{y}}{\frac{x}{y}+\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{y}{x}}{\frac{x}{y}+\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y}{\ln x}$
Explanation
This problem involves implicit differentiation for x ln y + y ln x = 10 to find dy/dx. Differentiating gives ln y + x (1/y) dy/dx + ln x dy/dx + y (1/x) = 0, with product rules. Dy/dx terms are (x/y) dy/dx + ln x dy/dx, factoring to (x/y + ln x) dy/dx. Constants are ln y + y/x, so dy/dx = - (ln y + y/x) / (x/y + ln x). Choice B tempts with x/y in numerator but fails by mismatching terms from product rule. Recognize this in symmetric logarithmic terms like x ln y and y ln x.
If $x^2\sin y + y^2\cos x=0$, what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{2x\sin y}{2y\cos x}$
$\dfrac{dy}{dx}=-\dfrac{2x\sin y-y^2\sin x}{x^2\cos y}$
$\dfrac{dy}{dx}=-\dfrac{2x\sin y-y^2\sin x}{x^2\cos y+2y\cos x}$
$\dfrac{dy}{dx}=-\dfrac{2x\sin y+y^2\sin x}{x^2\cos y+2y\cos x}$
$\dfrac{dy}{dx}=-\dfrac{2x\cos y-y^2\sin x}{x^2\cos y+2y\cos x}$
Explanation
This problem requires implicit differentiation for x² sin y + y² cos x = 0 to find dy/dx. Differentiating gives 2x sin y + x² cos y dy/dx + 2y cos x dy/dx - y² sin x = 0, using product and chain. Dy/dx terms are x² cos y dy/dx + 2y cos x dy/dx, grouping to (x² cos y + 2y cos x) dy/dx. Constants are 2x sin y - y² sin x, so dy/dx = - (2x sin y - y² sin x) / (x² cos y + 2y cos x), but A has + y² sin x in numerator. Wait, verifying: actually, moving terms, it's - (2x sin y + y² sin x wait no: constants 2x sin y - y² sin x to other side makes -(2x sin y - y² sin x) = -2x sin y + y² sin x, but A has 2x sin y - y² sin x in numerator with negative, which is equivalent after sign. No, A is - (2x sin y - y² sin x)/denom? Wait, A has - (2x sin y - y² sin x), but let's calculate properly. Differentiating: derivative of y² cos x is 2y dy/dx cos x + y² (-sin x), so 2y cos x dy/dx - y² sin x. So total: 2x sin y + x² cos y y' + 2y cos x y' - y² sin x = 0. So y' (x² cos y + 2y cos x) = -2x sin y + y² sin x. Thus y' = (-2x sin y + y² sin x) / (x² cos y + 2y cos x) = - (2x sin y - y² sin x) / denom, which is A. Yes. Choice B tempts by flipping signs in numerator but fails to match the correct collection. Recognize in zero-equated trig products.
If $\sin(xy)+x^2=y$, what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$?
$\dfrac{dy}{dx}=\dfrac{2x+\cos(xy)}{1-\cos(xy)}$
$\dfrac{dy}{dx}=\dfrac{2x+ y\cos(xy)}{1- x\cos(xy)}$
$\dfrac{dy}{dx}=2x+y\cos(xy)$
$\dfrac{dy}{dx}=\dfrac{2x}{1-x\cos(xy)}$
$\dfrac{dy}{dx}=\dfrac{2x+y\cos(xy)}{1+ x\cos(xy)}$
Explanation
This problem requires implicit differentiation to find dy/dx for the equation sin(xy) + x² = y. Differentiating both sides gives cos(xy) * (y + x dy/dx) + 2x = dy/dx, incorporating the chain rule for sin(xy). The dy/dx terms appear as cos(xy) * x dy/dx on the left and -dy/dx on the right, which can be grouped as [cos(xy) x - 1] dy/dx. Rearranging yields dy/dx = (2x + y cos(xy)) / (1 - x cos(xy)), with numerator from x terms and denominator from y terms. Choice D is a tempting distractor but fails due to a sign error in the denominator, using +x cos(xy) instead of -x cos(xy) from moving terms. To recognize implicit differentiation needs, identify equations mixing trig functions of products like xy with polynomials.
Suppose $x^2+\sin y= y\cos x$; what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=\dfrac{-2x}{\cos y}$
$\dfrac{dy}{dx}=\dfrac{-2x-y\sin x}{\cos y+\cos x}$
$\dfrac{dy}{dx}=\dfrac{-2x-y\sin x}{\cos y-\cos x}$
$\dfrac{dy}{dx}=\dfrac{-2x+y\sin x}{\cos y-\cos x}$
$\dfrac{dy}{dx}=-2x-y\sin x$
Explanation
This problem requires implicit differentiation for x² + sin y = y cos x to find dy/dx. Differentiating gives 2x + cos y dy/dx = dy/dx cos x - y sin x, using chain on sin y and product on y cos x. Dy/dx terms are cos y dy/dx - cos x dy/dx, grouping to (cos y - cos x) dy/dx. Constants are 2x + y sin x, so dy/dx = -(2x + y sin x)/(cos y - cos x), adjusting signs. Choice C tempts with +cos x in denominator but fails due to incorrect sign when collecting terms. Spot implicit differentiation in trig equations with x and y arguments.