We need to find $\frac{dy}{dx}$ for the given implicit equations.

1. Problem Description

We need to find $\frac{dy}{dx}$ for the given implicit equations.

2. Solution Steps

2

1. $x^3 + 2x^2y - y^3 = 0$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(2x^2y) - \frac{d}{dx}(y^3) = \frac{d}{dx}(0)$

$3x^2 + 2\frac{d}{dx}(x^2y) - 3y^2\frac{dy}{dx} = 0$

$3x^2 + 2(x^2\frac{dy}{dx} + y\frac{d}{dx}(x^2)) - 3y^2\frac{dy}{dx} = 0$

$3x^2 + 2(x^2\frac{dy}{dx} + y(2x)) - 3y^2\frac{dy}{dx} = 0$

$3x^2 + 2x^2\frac{dy}{dx} + 4xy - 3y^2\frac{dy}{dx} = 0$

$(2x^2 - 3y^2)\frac{dy}{dx} = -3x^2 - 4xy$

$\frac{dy}{dx} = \frac{-3x^2 - 4xy}{2x^2 - 3y^2} = \frac{3x^2 + 4xy}{3y^2 - 2x^2}$

2

2. $ye^{-x} + 5x - 17 = 0$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(ye^{-x}) + \frac{d}{dx}(5x) - \frac{d}{dx}(17) = \frac{d}{dx}(0)$

$\frac{d}{dx}(ye^{-x}) + 5 - 0 = 0$

$y\frac{d}{dx}(e^{-x}) + e^{-x}\frac{dy}{dx} + 5 = 0$

$y(-e^{-x}) + e^{-x}\frac{dy}{dx} + 5 = 0$

$-ye^{-x} + e^{-x}\frac{dy}{dx} = -5$

$e^{-x}\frac{dy}{dx} = ye^{-x} - 5$

$\frac{dy}{dx} = \frac{ye^{-x} - 5}{e^{-x}} = y - 5e^x$

2

3. $x\sin y + y\cos x = 0$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x\sin y) + \frac{d}{dx}(y\cos x) = \frac{d}{dx}(0)$

$x\frac{d}{dx}(\sin y) + \sin y\frac{d}{dx}(x) + y\frac{d}{dx}(\cos x) + \cos x\frac{dy}{dx} = 0$

$x(\cos y)\frac{dy}{dx} + \sin y(1) + y(-\sin x) + \cos x\frac{dy}{dx} = 0$

$x\cos y \frac{dy}{dx} + \sin y - y\sin x + \cos x \frac{dy}{dx} = 0$

$(x\cos y + \cos x)\frac{dy}{dx} = y\sin x - \sin y$

$\frac{dy}{dx} = \frac{y\sin x - \sin y}{x\cos y + \cos x}$

2

4. $x^2\cos y - y^2\sin x = 0$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^2\cos y) - \frac{d}{dx}(y^2\sin x) = \frac{d}{dx}(0)$

$(x^2\frac{d}{dx}(\cos y) + \cos y\frac{d}{dx}(x^2)) - (y^2\frac{d}{dx}(\sin x) + \sin x\frac{d}{dx}(y^2)) = 0$

$(x^2(-\sin y)\frac{dy}{dx} + \cos y(2x)) - (y^2\cos x + \sin x(2y\frac{dy}{dx})) = 0$

$-x^2\sin y \frac{dy}{dx} + 2x\cos y - y^2\cos x - 2y\sin x \frac{dy}{dx} = 0$

$(-x^2\sin y - 2y\sin x)\frac{dy}{dx} = y^2\cos x - 2x\cos y$

$\frac{dy}{dx} = \frac{y^2\cos x - 2x\cos y}{-x^2\sin y - 2y\sin x} = \frac{2x\cos y - y^2\cos x}{x^2\sin y + 2y\sin x}$

3. Final Answer

2

1. $\frac{dy}{dx} = \frac{3x^2 + 4xy}{3y^2 - 2x^2}$

2

2. $\frac{dy}{dx} = y - 5e^x$

2

3. $\frac{dy}{dx} = \frac{y\sin x - \sin y}{x\cos y + \cos x}$

2

4. $\frac{dy}{dx} = \frac{2x\cos y - y^2\cos x}{x^2\sin y + 2y\sin x}$