If -5xy^(2)-4y=4+x then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate Equation: Differentiate both sides of the equation with respect to x. We will use the product rule for differentiating -5xy^2 and the chain rule for differentiating -4y, since y is a function of x. d/dx(-5xy^2) + d/dx(-4y) = d/dx(4+x)Apply Product Rule: Apply the product rule to the term -5xy^2. The product rule states that d/dx(uv) = u(dv/dx) + v(du/dx), where u = -5y^2 and v = x. d/dx(-5xy^2) = -5y^2 * d/dx(x) + x * d/dx(-5y^2)Differentiate Terms: Differentiate x and -5y^2 with respect to x. d/dx(x) = 1 d/dx(-5y^2) = -5 * 2y * dy/dx (using the chain rule, since y is a function of x)Substitute Derivatives: Substitute the derivatives back into the equation. d/dx(-5xy^2) = -5y^2 * 1 + x * (-10y * dy/dx) d/dx(-4y) = -4 * dy/dx (using the chain rule)Differentiate Right Side: Differentiate the right side of the equation. d/dx(4+x) = 0 + 1Combine Differentiated Parts: Combine all the differentiated parts. -5y^2 - 10xy * dy/dx - 4 * dy/dx = 1Isolate Terms: Isolate terms with dy/dx on one side and move the rest to the other side. -10xy * dy/dx - 4 * dy/dx = 1 + 5y^2Factor Out dy/dx: Factor out dy/dx from the left side of the equation. dy/dx * (-10xy - 4) = 1 + 5y^2Solve for dy/dx: Solve for dy/dx. dy/dx = (1 + 5y^2) / (-10xy - 4)

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If
-5xy^(2)-4y=4+x then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-5 x y^{2}-4 y=4+x$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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Algebra 1

Transformations of quadratic functions

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Q. If

-5 x y^{2}-4 y=4+x

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate Equation: Differentiate both sides of the equation with respect to $x$ . We will use the product rule for differentiating $-5xy^2$ and the chain rule for differentiating $-4y$ , since $y$ is a function of $x$ . $\frac{d}{dx}(-5xy^2) + \frac{d}{dx}(-4y) = \frac{d}{dx}(4+x)$
Apply Product Rule: Apply the product rule to the term $-5xy^2$ . The product rule states that $\frac{d}{dx}(uv) = u\left(\frac{dv}{dx}\right) + v\left(\frac{du}{dx}\right)$ , where $u = -5y^2$ and $v = x$ . $\frac{d}{dx}(-5xy^2) = -5y^2 \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(-5y^2)$
Differentiate Terms: Differentiate $x$ and $-5y^2$ with respect to $x$ .
$\frac{d}{dx}(x) = 1$
$\frac{d}{dx}(-5y^2) = -5 \times 2y \times \frac{dy}{dx}$ (using the chain rule, since $y$ is a function of $x$ )
Substitute Derivatives: Substitute the derivatives back into the equation. $\newline$ $\frac{d}{dx}(-5xy^2) = -5y^2 \cdot 1 + x \cdot (-10y \cdot \frac{dy}{dx})$ $\newline$ $\frac{d}{dx}(-4y) = -4 \cdot \frac{dy}{dx}$ (using the chain rule)
Differentiate Right Side: Differentiate the right side of the equation. $\frac{d}{dx}(4+x) = 0 + 1$
Combine Differentiated Parts: Combine all the differentiated parts. $\newline$ $-5y^2 - 10xy \frac{dy}{dx} - 4 \frac{dy}{dx} = 1$
Isolate Terms: Isolate terms with $\frac{dy}{dx}$ on one side and move the rest to the other side. $\newline$ $-10xy \cdot \frac{dy}{dx} - 4 \cdot \frac{dy}{dx} = 1 + 5y^2$
Factor Out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation. $\newline$ $\frac{dy}{dx} \times (-10xy - 4) = 1 + 5y^2$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . $\frac{dy}{dx} = \frac{1 + 5y^2}{-10xy - 4}$