find the slope of the tangent to the curve ( x^{5}+2xy + y^{2}=64 ) at ( (1,7) ).\nthe slope is\n(enter…

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (x^{5}+2xy + y^{2}) term - by - term. Using the power rule ((x^{n})^\prime=nx^{n - 1}), ((uv)^\prime = u^\prime v+uv^\prime) (where (u = 2x), (v = y)) and ((y^{2})^\prime=2y\frac{dy}{dx}) (by the chain rule). The derivative of (x^{5}) is (5x^{4}), the derivative of (2xy) is (2y + 2x\frac{dy}{dx}), and the derivative of (y^{2}) is (2y\frac{dy}{dx}). The derivative of the constant (64) is (0). So, (5x^{4}+2y + 2x\frac{dy}{dx}+2y\frac{dy}{dx}=0).

Step2: Solve for (\frac{dy}{dx})

Group the terms with (\frac{dy}{dx}): (2x\frac{dy}{dx}+2y\frac{dy}{dx}=-5x^{4}-2y). Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(2x + 2y)=-5x^{4}-2y). Then (\frac{dy}{dx}=\frac{-5x^{4}-2y}{2x + 2y}).

Step3: Substitute (x = 1) and (y = 7)

(\frac{dy}{dx}\mid_{x = 1,y = 7}=\frac{-5(1)^{4}-2(7)}{2(1)+2(7)}). First, calculate the numerator: (-5\times1-2\times7=-5 - 14=-19). Then, calculate the denominator: (2\times1+2\times7=2 + 14 = 16).

Answer:

(-\frac{19}{16})