Question

find the slope of the tangent to the curve ( x^{5}+2xy + y^{2}=64 ) at ( (1,7) ).
the slope is
(enter undef if the slope is not defined at this point.)

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:

\(-\frac{19}{16}\)