If -4+5x^(3)=y^(3)+4x^(2)y then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate with respect to x: Differentiate both sides of the equation with respect to x. We will use the power rule for differentiation and the product rule for the term 4x^2y. Differentiate the left side: d/dx(-4 + 5x^3) = 0 + 15x^2 Differentiate the right side using the chain rule for y^3 and the product rule for 4x^2y: d/dx(y^3 + 4x^2y) = 3y^2(dy/dx) + 4(2xy + x^2(dy/dx))Set derivatives equal: Set the derivatives from both sides equal to each other. 15x^2 = 3y^2(dy/dx) + 4(2xy + x^2(dy/dx))Combine like terms: Distribute and combine like terms. 15x^2 = 3y^2(dy/dx) + 8xy + 4x^2(dy/dx)Isolate terms: Isolate terms with dy/dx on one side and the rest on the other side. 15x^2 - 8xy = 3y^2(dy/dx) + 4x^2(dy/dx)Factor out dy/dx: Factor out dy/dx from the right side of the equation. 15x^2 - 8xy = dy/dx(3y^2 + 4x^2)Solve for dy/dx: Solve for dy/dx by dividing both sides by (3y^2 + 4x^2). (dy/dx) = (15x^2 - 8xy) / (3y^2 + 4x^2)

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If
-4+5x^(3)=y^(3)+4x^(2)y then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-4+5 x^{3}=y^{3}+4 x^{2} y$ 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

Full solution

Q. If

-4+5 x^{3}=y^{3}+4 x^{2} y

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate with respect to $x$ : Differentiate both sides of the equation with respect to $x$ . We will use the power rule for differentiation and the product rule for the term $4x^2y$ . Differentiate the left side: $\frac{d}{dx}(-4 + 5x^3) = 0 + 15x^2$ Differentiate the right side using the chain rule for $y^3$ and the product rule for $4x^2y$ : $\frac{d}{dx}(y^3 + 4x^2y) = 3y^2\frac{dy}{dx} + 4(2xy + x^2\frac{dy}{dx})$
Set derivatives equal: Set the derivatives from both sides equal to each other. $\newline$ $15x^2 = 3y^2\frac{dy}{dx} + 4(2xy + x^2\frac{dy}{dx})$
Combine like terms: Distribute and combine like terms. $15x^2 = 3y^2\frac{dy}{dx} + 8xy + 4x^2\frac{dy}{dx}$
Isolate terms: Isolate terms with $\frac{dy}{dx}$ on one side and the rest on the other side. $15x^2 - 8xy = 3y^2\left(\frac{dy}{dx}\right) + 4x^2\left(\frac{dy}{dx}\right)$
Factor out dy/dx: Factor out $\frac{dy}{dx}$ from the right side of the equation. $\newline$ $15x^2 - 8xy = \frac{dy}{dx}(3y^2 + 4x^2)$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ by dividing both sides by $(3y^2 + 4x^2)$ . $\frac{dy}{dx} = \frac{15x^2 - 8xy}{3y^2 + 4x^2}$