Math / Calculus

QuestionIf $-y^{3}-3-4 x y+4 x=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ .

Studdy Solution

STEP 1

What is this asking? We're given an equation mixing $x$ and $y$ and we need to find the derivative of $y$ with respect to $x$ , written as $\frac{dy}{dx}$ . Watch out! Remember that $y$ is secretly a function of $x$ , so we'll need to use the chain rule when differentiating terms involving $y$ .
Don't treat $y$ as a constant!

STEP 2

1. Differentiate both sides
2. Isolate and solve for $\frac{dy}{dx}$

STEP 3

Let's rewrite the equation for clarity: $-y^3 - 3 - 4xy + 4x = 0$

STEP 4

Now, we differentiate both sides of the equation with respect to $x$ .
Remember, $y$ is a function of $x$ , so we'll need the chain rule! $\frac{d}{dx}(-y^3 - 3 - 4xy + 4x) = \frac{d}{dx}(0)$

STEP 5

$\frac{d}{dx}(-y^3) + \frac{d}{dx}(-3) + \frac{d}{dx}(-4xy) + \frac{d}{dx}(4x) = 0$ $-3y^2 \cdot \frac{dy}{dx} - 0 - 4\left(x \cdot \frac{dy}{dx} + y \cdot 1\right) + 4 = 0$ $-3y^2 \cdot \frac{dy}{dx} - 4x \cdot \frac{dy}{dx} - 4y + 4 = 0$

STEP 6

Let's group all the terms with $\frac{dy}{dx}$ together to isolate it: $-3y^2 \cdot \frac{dy}{dx} - 4x \cdot \frac{dy}{dx} = 4y - 4$ $\frac{dy}{dx}(-3y^2 - 4x) = 4y - 4$

STEP 7

Finally, we divide both sides by $(-3y^2 - 4x)$ to solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{4y - 4}{-3y^2 - 4x}$ $\frac{dy}{dx} = \frac{4(y - 1)}{-3y^2 - 4x}$

STEP 8

So, we found that $\frac{dy}{dx} = \frac{4(y - 1)}{-3y^2 - 4x}$ .
Awesome!

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