Math  /  Calculus

QuestionFind dy/dxd y / d x in terms of xx and yy if x3yx4y2=0x^{3} y-x-4 y-2=0. dydx=\frac{d y}{d x}= \square

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of yy with respect to xx, written as dydx\frac{dy}{dx}, given the equation x3yx4y2=0x^3 y - x - 4y - 2 = 0. Watch out! Remember that yy is a function of xx, so we'll need to use the **product rule** and **implicit differentiation**!
Don't accidentally treat yy as a constant.

STEP 2

1. Implicit Differentiation
2. Isolate and Simplify

STEP 3

Let's **differentiate** both sides of the equation x3yx4y2=0x^3 y - x - 4y - 2 = 0 with respect to xx.
Remember, we're thinking of yy as a function of xx, so we'll need the **chain rule** and the **product rule**.

STEP 4

ddx(x3y)=x3dydx+yddx(x3)=x3dydx+y3x2=x3dydx+3x2y\frac{d}{dx}(x^3 y) = x^3 \cdot \frac{dy}{dx} + y \cdot \frac{d}{dx}(x^3) = x^3 \cdot \frac{dy}{dx} + y \cdot 3x^2 = x^3 \frac{dy}{dx} + 3x^2y Here, we used the **product rule** on the first term.

STEP 5

ddx(x)=1\frac{d}{dx}(-x) = -1 ddx(4y)=4dydx\frac{d}{dx}(-4y) = -4 \frac{dy}{dx} ddx(2)=0\frac{d}{dx}(-2) = 0We differentiated the remaining terms with respect to xx.
For 4y-4y, we used the **chain rule**.

STEP 6

Putting it all together, we get: x3dydx+3x2y14dydx=0x^3 \frac{dy}{dx} + 3x^2 y - 1 - 4 \frac{dy}{dx} = 0 This is the result of differentiating both sides of our original equation.

STEP 7

Now, let's **isolate** dydx\frac{dy}{dx}.
We'll group all the terms with dydx\frac{dy}{dx} on one side: x3dydx4dydx=13x2yx^3 \frac{dy}{dx} - 4 \frac{dy}{dx} = 1 - 3x^2 y

STEP 8

We can **factor out** dydx\frac{dy}{dx}: dydx(x34)=13x2y\frac{dy}{dx}(x^3 - 4) = 1 - 3x^2 y

STEP 9

Finally, we can **solve** for dydx\frac{dy}{dx} by dividing both sides by (x34)(x^3 - 4): dydx=13x2yx34\frac{dy}{dx} = \frac{1 - 3x^2 y}{x^3 - 4} And there we have it!

STEP 10

dydx=13x2yx34\frac{dy}{dx} = \frac{1 - 3x^2 y}{x^3 - 4}

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