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 \( = \)

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: \[ \text{slope} = \frac{291}{47} \]