Pregunta
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Use implicit differentiation to find the slope of the tangent line to the curve \( \frac{y}{x+8 y}=x^{9}+5 \) at the point \( \left(1, \frac{6}{-47}\right) \) slope \( = \)

Ask by Little Bryant. in the United States
Mar 12,2025

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The slope of the tangent line is \( -\frac{273}{2209} \).

Solución

Find the derivative with respect to \( x \) for \( \frac{y}{x+8y}=x^{9}+5 \). Find the first derivative by following steps: - step0: Find the derivative with respect to \(x\): \(\frac{y}{x+8y}=x^{9}+5\) - step1: Take the derivative: \(\frac{d}{dx}\left(\frac{y}{x+8y}\right)=\frac{d}{dx}\left(x^{9}+5\right)\) - step2: Calculate the derivative: \(\frac{x\frac{dy}{dx}-y}{\left(x+8y\right)^{2}}=\frac{d}{dx}\left(x^{9}+5\right)\) - step3: Calculate the derivative: \(\frac{x\frac{dy}{dx}-y}{\left(x+8y\right)^{2}}=9x^{8}\) - step4: Cross multiply: \(x\frac{dy}{dx}-y=\left(x+8y\right)^{2}\times 9x^{8}\) - step5: Simplify the equation: \(x\frac{dy}{dx}-y=9\left(x+8y\right)^{2}x^{8}\) - step6: Move the constant to the right side: \(x\frac{dy}{dx}=9\left(x+8y\right)^{2}x^{8}+y\) - step7: Divide both sides: \(\frac{x\frac{dy}{dx}}{x}=\frac{9\left(x+8y\right)^{2}x^{8}+y}{x}\) - step8: Divide the numbers: \(\frac{dy}{dx}=\frac{9\left(x+8y\right)^{2}x^{8}+y}{x}\) - step9: Expand the expression: \(\frac{dy}{dx}=\frac{9x^{10}+144x^{9}y+576y^{2}x^{8}+y}{x}\) Substitute \( x=1 \) and \( y=\frac{6}{-47} \) into the derivative \( \frac{dy}{dx}=\frac{9x^{10}+144x^{9}y+576y^{2}x^{8}+y}{x} \). Evaluate the expression by following steps: - step0: Evaluate: \(\frac{9x^{10}+144x^{9}y+576y^{2}x^{8}+y}{x}\) - step1: Substitute: \(\frac{9\times 1^{10}+144\times 1^{9}\times \frac{6}{-47}+576\left(\frac{6}{-47}\right)^{2}\times 1^{8}+\frac{6}{-47}}{1}\) - step2: Rewrite the fraction: \(\frac{9\times 1^{10}+144\times 1^{9}\times \frac{6}{-47}+576\left(-\frac{6}{47}\right)^{2}\times 1^{8}+\frac{6}{-47}}{1}\) - step3: Evaluate the power: \(\frac{9\times 1+144\times 1^{9}\times \frac{6}{-47}+576\left(-\frac{6}{47}\right)^{2}\times 1^{8}+\frac{6}{-47}}{1}\) - step4: Evaluate the power: \(\frac{9\times 1+144\times 1\times \frac{6}{-47}+576\left(-\frac{6}{47}\right)^{2}\times 1^{8}+\frac{6}{-47}}{1}\) - step5: Rewrite the fraction: \(\frac{9\times 1+144\times 1\times \left(-\frac{6}{47}\right)+576\left(-\frac{6}{47}\right)^{2}\times 1^{8}+\frac{6}{-47}}{1}\) - step6: Evaluate the power: \(\frac{9\times 1+144\times 1\times \left(-\frac{6}{47}\right)+576\left(-\frac{6}{47}\right)^{2}\times 1+\frac{6}{-47}}{1}\) - step7: Multiply: \(\frac{9+144\times 1\times \left(-\frac{6}{47}\right)+576\left(-\frac{6}{47}\right)^{2}\times 1+\frac{6}{-47}}{1}\) - step8: Multiply the terms: \(\frac{9-\frac{864}{47}+576\left(-\frac{6}{47}\right)^{2}\times 1+\frac{6}{-47}}{1}\) - step9: Multiply the terms: \(\frac{9-\frac{864}{47}+\frac{20736}{2209}+\frac{6}{-47}}{1}\) - step10: Rewrite the fraction: \(\frac{9-\frac{864}{47}+\frac{20736}{2209}-\frac{6}{47}}{1}\) - step11: Calculate: \(\frac{-\frac{273}{2209}}{1}\) - step12: Divide the terms: \(-\frac{273}{2209}\) The slope of the tangent line to the curve \( \frac{y}{x+8y}=x^{9}+5 \) at the point \( \left(1, \frac{6}{-47}\right) \) is \( -\frac{273}{2209} \).

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To find the slope of the tangent line using implicit differentiation, we start from the equation: \[ \frac{y}{x + 8y} = x^9 + 5 \] First, we can cross-multiply to eliminate the fraction: \[ y = (x + 8y)(x^9 + 5) \] Next, we will differentiate both sides with respect to \(x\). Using the product rule on the right side and remembering to apply the chain rule where needed, we get: \[ \frac{dy}{dx} = (x + 8y)'(x^9 + 5) + (x + 8y)(x^9 + 5)' \] Calculating this gives us: \[ \frac{dy}{dx} = (1 + 8 \frac{dy}{dx})(x^9 + 5) + (x + 8y)(9x^8) \] Now we can isolate \(\frac{dy}{dx}\). Group all \(\frac{dy}{dx}\) terms on one side: \[ \frac{dy}{dx} = (1 + 8 \frac{dy}{dx})(x^9 + 5) + (x + 8y)(9x^8) \] This simplifies to: \[ \frac{dy}{dx} - 8(x^9 + 5) \frac{dy}{dx} = (x + 8y)(9x^8) - (x^9 + 5) \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(1 - 8(x^9 + 5)) = (x + 8y)(9x^8) - (x^9 + 5) \] Now we can solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{(x + 8y)(9x^8) - (x^9 + 5)}{1 - 8(x^9 + 5)} \] Now we substitute \(x = 1\) and \(y = \frac{6}{-47}\): Substituting \(x = 1\) and \(y = \frac{6}{-47}\): \[ \frac{dy}{dx} = \frac{(1 + 8 \cdot \frac{6}{-47})(9 \cdot 1^8) - (1^9 + 5)}{1 - 8(1^9 + 5)} \] Calculating this gives: The numerator: \[ (1 + \frac{48}{-47}) \cdot 9 - 6 = \frac{-47 + 48}{-47} \cdot 9 - 6 = \frac{1}{-47} \cdot 9 - 6 \] The denominator: \[ 1 - 8(6) = 1 - 48 = -47 \] Putting it all together: The slope \(\frac{dy}{dx}\) simplifies to: \[ \frac{(-9/47 - 6)}{-47} \] This calculation leads to the final slope \(=\frac{9 + 282}{47} = \frac{291}{47}\). So the slope of the tangent line at the given point is: \[ \text{slope} = \frac{291}{47} \]

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