Implicit Differentiation
Help Questions
AP Calculus AB › Implicit Differentiation
For the level curve $e^{x}+e^{y}=xy$, what is $\dfrac{dy}{dx}$ at a general point?
$\dfrac{dy}{dx}=\dfrac{e^{x}}{x}$
$\dfrac{dy}{dx}=\dfrac{y-e^{x}}{e^{y}-x}$
$\dfrac{dy}{dx}=\dfrac{y}{e^{y}}$
$\dfrac{dy}{dx}=\dfrac{y-e^{x}}{e^{y}+x}$
$\dfrac{dy}{dx}=\dfrac{y+e^{x}}{e^{y}+x}$
Explanation
This problem uses implicit differentiation for the level curve $e^x$ + $e^y$ = xy to find dy/dx. Differentiating yields $e^x$ + $e^y$ dy/dx = y + x dy/dx, with chain rule on $e^y$ and product rule on xy. Dy/dx terms are $e^y$ dy/dx - x dy/dx, grouping to $(e^y$ - x) dy/dx, and constants y - $e^x$. Thus, dy/dx = (y - $e^x$$)/(e^y$ - x), noting the negative in numerator for solving. Choice C is tempting but fails by using +x in denominator, likely forgetting to move x dy/dx correctly. Spot implicit differentiation in equations with exponentials of x and y equated to products.
For the relation $x\ln y+y\ln x=10$, what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{x}{y}}{\frac{x}{y}+\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{y}{x}}{\frac{x}{y}+\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y+\frac{y}{x}}{\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\ln y}{\ln x}$
$\dfrac{dy}{dx}=\dfrac{\ln y+\frac{y}{x}}{\frac{x}{y}+\ln x}$
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.
For the curve defined by $x^2+xy+y^2=7$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{1+2y}$
$\dfrac{dy}{dx}=-\dfrac{2x+xy+y^2}{x^2+x+2y}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+y}$
$\dfrac{dy}{dx}=-\dfrac{2x}{x}$
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.
If $\sqrt{x^2+y^2}=x+y$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=\dfrac{x-1}{1-y}$
$\dfrac{dy}{dx}=\dfrac{\frac{x}{\sqrt{x^2+y^2}}-1}{1+\frac{y}{\sqrt{x^2+y^2}}}$
$\dfrac{dy}{dx}=\dfrac{1-\frac{x}{\sqrt{x^2+y^2}}}{1-\frac{y}{\sqrt{x^2+y^2}}}$
$\dfrac{dy}{dx}=\dfrac{\frac{x}{\sqrt{x^2+y^2}}-1}{1-\frac{y}{\sqrt{x^2+y^2}}}$
$\dfrac{dy}{dx}=\dfrac{\frac{x}{\sqrt{x^2+y^2}}}{\frac{y}{\sqrt{x^2+y^2}}}$
Explanation
This problem requires implicit differentiation for $\sqrt{x^2 + y^2} = x + y$ to find $\frac{dy}{dx}$. Differentiating gives $\frac{1}{2\sqrt{x^2 + y^2}} (2x + 2y \frac{dy}{dx}) = 1 + \frac{dy}{dx}$, chain rule on left. Simplifies to $\frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}} = 1 + \frac{dy}{dx}$. Dy/dx terms: $[ \frac{y}{\sqrt{x^2 + y^2}} - 1 ] \frac{dy}{dx} = 1 - \frac{x}{\sqrt{x^2 + y^2}} $, but A has numerator $\frac{x}{\sqrt{x^2 + y^2}} - 1$ over $1 - \frac{y}{\sqrt{x^2 + y^2}}$. Yes, matching after rearrangement. Choice B tempts with + in denominator but fails by not accounting for the sign when isolating dy/dx. Recognize in distance-like equations with square roots of sums.
If $\ln(x+y)=xy$, what is $\dfrac{dy}{dx}$ at a general point $(x,y)$?
$\dfrac{dy}{dx}=\dfrac{1}{x+y}$
$\dfrac{dy}{dx}=\dfrac{y(x+y)-1}{1+x(x+y)}$
$\dfrac{dy}{dx}=y$
$\dfrac{dy}{dx}=\dfrac{y(x+y)-1}{1-x(x+y)}$
$\dfrac{dy}{dx}=\dfrac{1-y(x+y)}{1-x(x+y)}$
Explanation
This problem requires implicit differentiation for ln(x + y) = xy to find dy/dx. Differentiating gives [1/(x + y)] (1 + dy/dx) = y + x dy/dx, applying chain rule to ln and product to xy. Dy/dx terms are [1/(x + y)] dy/dx - x dy/dx, grouping to {1/(x + y) - x} dy/dx. Constants are y - 1/(x + y), so dy/dx = [y (x + y) - 1]/[1 - x (x + y)] after multiplying numerator and denominator by x + y. Choice B is tempting but fails with a sign error in the numerator, using 1 - y(x + y) instead. Identify implicit differentiation in logarithmic equations involving sums like x + y.
If $y^2\ln x + x\ln y = 3$, what is $\dfrac{dy}{dx}$ at a general point?
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}+\ln y}{2\ln x+\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}}{2y\ln x}$
$\dfrac{dy}{dx}=-\dfrac{\frac{2y}{x}+\ln y}{2y\ln x+\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y^2}{x}+\ln y}{2y\ln x+\frac{x}{y}}$
$\dfrac{dy}{dx}=-\dfrac{\ln y}{\frac{x}{y}}$
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.
If $y,\sqrt{x}+x,\sqrt{y}=6$, what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=-\dfrac{\frac{1}{2\sqrt{x}}+\sqrt{y}}{\sqrt{x}+\frac{x}{2\sqrt{y}}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y}{2\sqrt{x}}+\sqrt{y}}{\sqrt{x}-\frac{x}{2\sqrt{y}}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y}{2\sqrt{x}}}{\sqrt{x}}$
$\dfrac{dy}{dx}=\dfrac{\frac{y}{2\sqrt{x}}+\sqrt{y}}{\sqrt{x}+\frac{x}{2\sqrt{y}}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{y}{2\sqrt{x}}+\sqrt{y}}{\sqrt{x}+\frac{x}{2\sqrt{y}}}$
Explanation
This problem requires implicit differentiation for $y \sqrt{x} + x \sqrt{y} = 6$ to find $\dfrac{dy}{dx}$. Differentiating gives $y \left( \frac{1}{2 \sqrt{x}} \right) + \sqrt{x} \dfrac{dy}{dx} + \sqrt{y} + x \left( \frac{1}{2 \sqrt{y}} \right) \dfrac{dy}{dx} = 0$, product and chain. $\dfrac{dy}{dx}$ terms: $\sqrt{x} \dfrac{dy}{dx} + \left( \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx}$, grouping to $\left( \sqrt{x} + \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx}$. Constants: $\frac{y}{2 \sqrt{x}} + \sqrt{y}$, so $\dfrac{dy}{dx} = - \left( \frac{y}{2 \sqrt{x}} + \sqrt{y} \right) / \left( \sqrt{x} + \frac{x}{2 \sqrt{y}} \right)$. Choice B tempts with $-$ in denominator but fails by changing sign incorrectly. Recognize in products of variables and their roots.
For the relation $x^2y+xy^2=8$, what is $\dfrac{dy}{dx}$ at $(x,y)$?
$\dfrac{dy}{dx}=\dfrac{2xy+y^2}{x^2+2xy}$
$\dfrac{dy}{dx}=-\dfrac{2xy+y^2}{x^2+xy}$
$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$
$\dfrac{dy}{dx}=-\dfrac{2xy+y^2}{x^2+2xy}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+2xy}$
Explanation
This problem involves implicit differentiation for x² y + x y² = 8 to find dy/dx. Differentiating gives 2x y + x² dy/dx + y² + 2 x y dy/dx = 0, product rules on both terms. Dy/dx terms: x² dy/dx + 2 x y dy/dx, factoring to (x² + 2 x y) dy/dx. Constants: 2x y + y², so dy/dx = - (2x y + y²) / (x² + 2 x y). Choice B tempts with x² + x y in denominator but fails by undercounting the coefficient from derivative of y². Recognize in factored forms like xy (x + y) = 8.
For $x^4+y^4=2x^2y^2+3$, what is $\dfrac{dy}{dx}$ at a general point?
$\dfrac{dy}{dx}=\dfrac{4x^3}{4y^3}$
$\dfrac{dy}{dx}=\dfrac{x(x^2-y^2)}{y(y^2+x^2)}$
$\dfrac{dy}{dx}=\dfrac{x(x^2-y^2)}{y(y^2-x^2)}$
$\dfrac{dy}{dx}=-\dfrac{x(x^2-y^2)}{y(y^2-x^2)}$
$\dfrac{dy}{dx}=\dfrac{x^3-y^3}{y^3-x^3}$
Explanation
This problem uses implicit differentiation for $x^4$ + $y^4$ = 2 x² y² + 3 to find dy/dx. Differentiating gives 4 x³ + 4 y³ dy/dx = 4 x y² + 2 x² (2 y dy/dx), product on right. Dy/dx terms: 4 y³ dy/dx - 4 x² y dy/dx, grouping to (4 y³ - 4 x² y) dy/dx. Constants: 4 x³ - 4 x y², so dy/dx = (4 x y² - 4 x³) / (4 y³ - 4 x² y) = - (x (x² - y²)) / (y (y² - x²)), matching D. Choice A tempts without negative but fails to account for sign when solving. Recognize in high-power symmetric equations.
For the curve $x^2y+\tan y=5$, what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$?
$\dfrac{dy}{dx}=-\dfrac{2x}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2}$
$\dfrac{dy}{dx}=\dfrac{2xy}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+\sec^2 y}$
$\dfrac{dy}{dx}=-\dfrac{2xy}{x^2-\sec^2 y}$
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.