Math / Calculus

QuestionFind $\frac{d y}{d x}$ by implicit differentiation. $x^{2}+4 x y-y^{3}=6$

Studdy Solution

STEP 1

1. The equation $x^2 + 4xy - y^3 = 6$ involves both $x$ and $y$ , and we need to find $\frac{dy}{dx}$ using implicit differentiation.
2. Both $x$ and $y$ are functions of each other, and $y$ is implicitly defined in terms of $x$ .

STEP 2

1. Differentiate both sides of the equation with respect to $x$ .
2. Solve for $\frac{dy}{dx}$ .

STEP 3

Differentiate each term of the equation $x^2 + 4xy - y^3 = 6$ with respect to $x$ .
- Differentiate $x^2$ with respect to $x$ : $ \frac{d}{dx}(x^2) = 2x \]
- Differentiate $4xy$ with respect to $x$ using the product rule: $ \frac{d}{dx}(4xy) = 4 \left( \frac{d}{dx}(x)y + x\frac{dy}{dx} \right) = 4(y + x\frac{dy}{dx}) \]
- Differentiate $-y^3$ with respect to $x$ using the chain rule: $ \frac{d}{dx}(-y^3) = -3y^2 \frac{dy}{dx} \]
- Differentiate the constant $6$ with respect to $x$ : $ \frac{d}{dx}(6) = 0 \]
Combine these results to form the differentiated equation: $2x + 4(y + x\frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0$

STEP 4

Solve for $\frac{dy}{dx}$ in the equation obtained from differentiation.
- Expand and simplify the differentiated equation: $ 2x + 4y + 4x\frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0 \]
- Collect all terms involving $\frac{dy}{dx}$ on one side: $ 4x\frac{dy}{dx} - 3y^2 \frac{dy}{dx} = -2x - 4y \]
- Factor out $\frac{dy}{dx}$ : $ \frac{dy}{dx}(4x - 3y^2) = -2x - 4y \]
- Solve for $\frac{dy}{dx}$ : $ \frac{dy}{dx} = \frac{-2x - 4y}{4x - 3y^2} \]
The derivative $\frac{dy}{dx}$ is:
$\frac{dy}{dx} = \frac{-2x - 4y}{4x - 3y^2}$

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