Exploring Behaviors of Implicit Relations - AP Calculus AB
Card 1 of 30
Find $\frac{dy}{dx}$ for $y = \ln(xy)$ implicitly.
Find $\frac{dy}{dx}$ for $y = \ln(xy)$ implicitly.
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$\frac{dy}{dx} = \frac{1}{x + y}$. From $\frac{dy}{dx} = \frac{1}{xy}(y + x\frac{dy}{dx})$, solve for $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{1}{x + y}$. From $\frac{dy}{dx} = \frac{1}{xy}(y + x\frac{dy}{dx})$, solve for $\frac{dy}{dx}$.
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Differentiate $x^2 - y^2 = 1$ implicitly.
Differentiate $x^2 - y^2 = 1$ implicitly.
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$2x - 2y \frac{dy}{dx} = 0$. Difference of squares: coefficients have opposite signs.
$2x - 2y \frac{dy}{dx} = 0$. Difference of squares: coefficients have opposite signs.
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Differentiate $x^3 + y^3 = 3xy$ implicitly.
Differentiate $x^3 + y^3 = 3xy$ implicitly.
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$3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \frac{dy}{dx})$. Apply product rule to right side.
$3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \frac{dy}{dx})$. Apply product rule to right side.
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Differentiate $x^2y^2 = 1$ implicitly.
Differentiate $x^2y^2 = 1$ implicitly.
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$2xy^2 + 2x^2y \frac{dy}{dx} = 0$. Product rule: $2xy^2 + 2x^2y\frac{dy}{dx} = 0$.
$2xy^2 + 2x^2y \frac{dy}{dx} = 0$. Product rule: $2xy^2 + 2x^2y\frac{dy}{dx} = 0$.
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Differentiate $x + y = e^{xy}$ implicitly.
Differentiate $x + y = e^{xy}$ implicitly.
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$1 + \frac{dy}{dx} = e^{xy}(y + x \frac{dy}{dx})$. Chain rule on exponential function on right side.
$1 + \frac{dy}{dx} = e^{xy}(y + x \frac{dy}{dx})$. Chain rule on exponential function on right side.
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Differentiate $y = x^2 + \tan(y)$ implicitly.
Differentiate $y = x^2 + \tan(y)$ implicitly.
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$\frac{dy}{dx} = 2x + \sec^2(y) \frac{dy}{dx}$. Chain rule on $\tan(y)$ gives $\sec^2(y)\frac{dy}{dx}$.
$\frac{dy}{dx} = 2x + \sec^2(y) \frac{dy}{dx}$. Chain rule on $\tan(y)$ gives $\sec^2(y)\frac{dy}{dx}$.
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Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.
Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.
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$\frac{dy}{dx} = \frac{y - 2x}{2y - x}$. Differentiate each term, collect $\frac{dy}{dx}$ terms, solve.
$\frac{dy}{dx} = \frac{y - 2x}{2y - x}$. Differentiate each term, collect $\frac{dy}{dx}$ terms, solve.
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Differentiate $x^4 + y^4 = 16$ implicitly.
Differentiate $x^4 + y^4 = 16$ implicitly.
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$4x^3 + 4y^3 \frac{dy}{dx} = 0$. Apply power rule: $4x^3 + 4y^3\frac{dy}{dx} = 0$.
$4x^3 + 4y^3 \frac{dy}{dx} = 0$. Apply power rule: $4x^3 + 4y^3\frac{dy}{dx} = 0$.
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Differentiate $x^2 + y^2 = 1$ implicitly with respect to $x$.
Differentiate $x^2 + y^2 = 1$ implicitly with respect to $x$.
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$2x + 2y \frac{dy}{dx} = 0$. Apply power rule to both terms, treat $y$ as function of $x$.
$2x + 2y \frac{dy}{dx} = 0$. Apply power rule to both terms, treat $y$ as function of $x$.
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What is the implicit differentiation of $x^2y + y^2 = 1$?
What is the implicit differentiation of $x^2y + y^2 = 1$?
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$2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$. Product rule on $x^2y$, power rule on $y^2$.
$2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$. Product rule on $x^2y$, power rule on $y^2$.
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What is the implicit differential of $x^2 + xy = 10$?
What is the implicit differential of $x^2 + xy = 10$?
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$2x + y + x \frac{dy}{dx} = 0$. Product rule on $xy$ term.
$2x + y + x \frac{dy}{dx} = 0$. Product rule on $xy$ term.
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What is the implicit derivative of $y^3 + 3x^2y = 12$?
What is the implicit derivative of $y^3 + 3x^2y = 12$?
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$3y^2 \frac{dy}{dx} + 6xy + 3x^2 \frac{dy}{dx} = 0$. Apply chain rule to $y^3$ and product rule to $3x^2y$.
$3y^2 \frac{dy}{dx} + 6xy + 3x^2 \frac{dy}{dx} = 0$. Apply chain rule to $y^3$ and product rule to $3x^2y$.
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Differentiate $y^2 = x + x^2y$ implicitly.
Differentiate $y^2 = x + x^2y$ implicitly.
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$2y \frac{dy}{dx} = 1 + 2xy + x^2 \frac{dy}{dx}$. Product rule on right side, power rule on left.
$2y \frac{dy}{dx} = 1 + 2xy + x^2 \frac{dy}{dx}$. Product rule on right side, power rule on left.
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Differentiate $\sin(x + y) = y$ implicitly.
Differentiate $\sin(x + y) = y$ implicitly.
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$\cos(x + y)(1 + \frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on $\sin(x + y)$ gives $\cos(x + y)(1 + \frac{dy}{dx})$.
$\cos(x + y)(1 + \frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on $\sin(x + y)$ gives $\cos(x + y)(1 + \frac{dy}{dx})$.
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What does $\frac{dy}{dx}$ represent in implicit differentiation?
What does $\frac{dy}{dx}$ represent in implicit differentiation?
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The derivative of $y$ with respect to $x$. The rate of change of $y$ with respect to $x$.
The derivative of $y$ with respect to $x$. The rate of change of $y$ with respect to $x$.
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What is the implicit derivative of $y = \sin(xy)$?
What is the implicit derivative of $y = \sin(xy)$?
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$\frac{dy}{dx} = \cos(xy)(y + x \frac{dy}{dx})$. Chain rule on $\sin(xy)$ equals $\frac{dy}{dx}$.
$\frac{dy}{dx} = \cos(xy)(y + x \frac{dy}{dx})$. Chain rule on $\sin(xy)$ equals $\frac{dy}{dx}$.
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Differentiate $x = \cos(xy)$ implicitly.
Differentiate $x = \cos(xy)$ implicitly.
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$1 = -\sin(xy)(y + x \frac{dy}{dx})$. Derivative of $\cos$ is $-\sin$, apply chain rule.
$1 = -\sin(xy)(y + x \frac{dy}{dx})$. Derivative of $\cos$ is $-\sin$, apply chain rule.
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Find $\frac{dy}{dx}$ for $y = \cos(xy)$ implicitly.
Find $\frac{dy}{dx}$ for $y = \cos(xy)$ implicitly.
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$\frac{dy}{dx} = -\sin(xy)(y + x \frac{dy}{dx})$. Chain rule: $\frac{dy}{dx} = -\sin(xy)(y + x\frac{dy}{dx})$.
$\frac{dy}{dx} = -\sin(xy)(y + x \frac{dy}{dx})$. Chain rule: $\frac{dy}{dx} = -\sin(xy)(y + x\frac{dy}{dx})$.
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What is the implicit derivative of $y = x + \sin(y)$?
What is the implicit derivative of $y = x + \sin(y)$?
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$\frac{dy}{dx} = 1 + \cos(y) \frac{dy}{dx}$. Chain rule on $\sin(y)$ gives $\cos(y)\frac{dy}{dx}$.
$\frac{dy}{dx} = 1 + \cos(y) \frac{dy}{dx}$. Chain rule on $\sin(y)$ gives $\cos(y)\frac{dy}{dx}$.
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Find $\frac{dy}{dx}$ for $xy + y = 3x$.
Find $\frac{dy}{dx}$ for $xy + y = 3x$.
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$\frac{dy}{dx} = \frac{3 - y}{x + 1}$. Factor out terms with $y$: $y(x + 1) = 3x$.
$\frac{dy}{dx} = \frac{3 - y}{x + 1}$. Factor out terms with $y$: $y(x + 1) = 3x$.
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Differentiate $x = \tan(xy)$ implicitly.
Differentiate $x = \tan(xy)$ implicitly.
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$1 = \sec^2(xy)(y + x \frac{dy}{dx})$. Derivative of $\tan$ is $\sec^2$, apply chain rule.
$1 = \sec^2(xy)(y + x \frac{dy}{dx})$. Derivative of $\tan$ is $\sec^2$, apply chain rule.
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Find $\frac{dy}{dx}$ for $xy = \ln(x)$.
Find $\frac{dy}{dx}$ for $xy = \ln(x)$.
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$\frac{dy}{dx} = \frac{1 - y}{x}$. From product rule: $y + x\frac{dy}{dx} = \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1 - y}{x}$. From product rule: $y + x\frac{dy}{dx} = \frac{1}{x}$.
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Differentiate $x^2 + y^2 = 25$ implicitly.
Differentiate $x^2 + y^2 = 25$ implicitly.
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$2x + 2y \frac{dy}{dx} = 0$. Same as $x^2 + y^2 = 1$ with different radius.
$2x + 2y \frac{dy}{dx} = 0$. Same as $x^2 + y^2 = 1$ with different radius.
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What is the implicit derivative of $x^3 + y^3 = 6xy$?
What is the implicit derivative of $x^3 + y^3 = 6xy$?
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$3x^2 + 3y^2 \frac{dy}{dx} = 6(y + x \frac{dy}{dx})$. Apply product rule to right side: $6(y + x\frac{dy}{dx})$.
$3x^2 + 3y^2 \frac{dy}{dx} = 6(y + x \frac{dy}{dx})$. Apply product rule to right side: $6(y + x\frac{dy}{dx})$.
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What is implicit differentiation?
What is implicit differentiation?
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Differentiating equations not solved for one variable in terms of others. Used when $y$ cannot be easily isolated.
Differentiating equations not solved for one variable in terms of others. Used when $y$ cannot be easily isolated.
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State the Chain Rule for implicit differentiation.
State the Chain Rule for implicit differentiation.
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Differentiate outer, multiply by derivative of inner. Essential for composite functions involving $y$.
Differentiate outer, multiply by derivative of inner. Essential for composite functions involving $y$.
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What is the purpose of implicit differentiation?
What is the purpose of implicit differentiation?
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To find derivatives when not in explicit form. Enables differentiation of relations not solved for $y$.
To find derivatives when not in explicit form. Enables differentiation of relations not solved for $y$.
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Differentiate $e^{xy} = x + y$ implicitly with respect to $x$.
Differentiate $e^{xy} = x + y$ implicitly with respect to $x$.
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$e^{xy}(y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx}$. Chain rule on left, standard derivatives on right.
$e^{xy}(y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx}$. Chain rule on left, standard derivatives on right.
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Differentiate $\tan(xy) = x$ implicitly.
Differentiate $\tan(xy) = x$ implicitly.
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$\sec^2(xy)(y + x \frac{dy}{dx}) = 1$. Chain rule: derivative of $\tan$ is $\sec^2$.
$\sec^2(xy)(y + x \frac{dy}{dx}) = 1$. Chain rule: derivative of $\tan$ is $\sec^2$.
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Find $\frac{dy}{dx}$ for $\cos(xy) = x$.
Find $\frac{dy}{dx}$ for $\cos(xy) = x$.
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$\frac{dy}{dx} = \frac{y\sin(xy) - 1}{x\sin(xy)}$. Chain rule gives $-\sin(xy)(y + x\frac{dy}{dx}) = 1$, solve.
$\frac{dy}{dx} = \frac{y\sin(xy) - 1}{x\sin(xy)}$. Chain rule gives $-\sin(xy)(y + x\frac{dy}{dx}) = 1$, solve.
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