use implicit differentiation to find the slope of the tangent line to the curve\n\\( \\frac { y } { x + 6 y…

Explanation:

Step1: Differentiate both sides with respect to (x)

Using the quotient rule (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}) for the left - hand side. Let (u = y) and (v=x + 6y). Then (u'=\frac{dy}{dx}=y') and (v'=1 + 6y'). The left - hand side derivative is (\frac{y'(x + 6y)-y(1 + 6y')}{(x + 6y)^{2}}). The right - hand side derivative is (8x^{7}) using the power rule ((x^{n})'=nx^{n - 1}). So we have (\frac{y'(x + 6y)-y(1 + 6y')}{(x + 6y)^{2}}=8x^{7}).

Step2: Expand the left - hand side

[ \begin{align*} \frac{y'x+6y'y - y-6y'y}{(x + 6y)^{2}}&=8x^{7}\ \frac{y'x - y}{(x + 6y)^{2}}&=8x^{7} \end{align*} ]

Step3: Solve for (y')

Multiply both sides by ((x + 6y)^{2}): (y'x - y=8x^{7}(x + 6y)^{2}). Then (y'x=8x^{7}(x + 6y)^{2}+y), and (y'=\frac{8x^{7}(x + 6y)^{2}+y}{x}).

Step4: Substitute (x = 1) and (y=\frac{3}{-17})

First, (x+6y=1+6\times\left(\frac{3}{-17}\right)=1-\frac{18}{17}=\frac{17 - 18}{17}=-\frac{1}{17}). Then (8x^{7}(x + 6y)^{2}+y=8\times1^{7}\times\left(-\frac{1}{17}\right)^{2}+\left(\frac{3}{-17}\right)=8\times\frac{1}{289}-\frac{3}{17}=\frac{8-51}{289}=-\frac{43}{289}). And (y'=\frac{-\frac{43}{289}}{1}=-\frac{43}{289}).

Answer:

(-\frac{43}{289})