Math  /  Calculus

QuestionFind dy/dxd y / d x by implicit differentiation x3y5+3x=8y3+1x^{3} y^{5}+3 x=8 y^{3}+1

Studdy Solution

STEP 1

1. The equation x3y5+3x=8y3+1x^{3} y^{5} + 3x = 8y^{3} + 1 involves both xx and yy.
2. We will use implicit differentiation to find dydx\frac{dy}{dx}.

STEP 2

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

STEP 3

Differentiate both sides of the equation with respect to xx. Remember to use the product rule and chain rule where necessary.
Differentiate the left side: - For x3y5x^3 y^5, use the product rule: \frac{d}{dx}(x^3 y^5) = x^3 \frac{d}{dx}(y^5) + y^5 \frac{d}{dx}(x^3) \] - Differentiate \(y^5\) with respect to \(x\): \frac{d}{dx}(y^5) = 5y^4 \frac{dy}{dx} \] - Differentiate x3x^3 with respect to xx: \frac{d}{dx}(x^3) = 3x^2 \] - Combine using the product rule: x^3 \cdot 5y^4 \frac{dy}{dx} + y^5 \cdot 3x^2 \]
- Differentiate 3x3x with respect to xx: $ \frac{d}{dx}(3x) = 3 \]
Differentiate the right side: - For 8y38y^3, use the chain rule: $ \frac{d}{dx}(8y^3) = 8 \cdot 3y^2 \frac{dy}{dx} = 24y^2 \frac{dy}{dx} \]
- Differentiate 11 with respect to xx: $ \frac{d}{dx}(1) = 0 \]
Combine the derivatives: x35y4dydx+y53x2+3=24y2dydxx^3 \cdot 5y^4 \frac{dy}{dx} + y^5 \cdot 3x^2 + 3 = 24y^2 \frac{dy}{dx}

STEP 4

Solve for dydx\frac{dy}{dx}.
Rearrange the equation to collect all terms involving dydx\frac{dy}{dx} on one side: x35y4dydx24y2dydx=y53x23x^3 \cdot 5y^4 \frac{dy}{dx} - 24y^2 \frac{dy}{dx} = - y^5 \cdot 3x^2 - 3
Factor out dydx\frac{dy}{dx}: dydx(x35y424y2)=y53x23\frac{dy}{dx} (x^3 \cdot 5y^4 - 24y^2) = - y^5 \cdot 3x^2 - 3
Solve for dydx\frac{dy}{dx}: dydx=y53x23x35y424y2\frac{dy}{dx} = \frac{- y^5 \cdot 3x^2 - 3}{x^3 \cdot 5y^4 - 24y^2}
The derivative dydx\frac{dy}{dx} is:
dydx=3x2y535x3y424y2\frac{dy}{dx} = \frac{-3x^2 y^5 - 3}{5x^3 y^4 - 24y^2}

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