use implicit differentiation to find an equation of the tangent line to the curve at the given point. 6x² +…

Explanation:

Step1: Differentiate both sides

Differentiate $6x^{2}+xy + 6y^{2}=13$ with respect to $x$. Using the sum - rule and product - rule: The derivative of $6x^{2}$ is $12x$. For $xy$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v = y$, we get $y+xy^\prime$. The derivative of $6y^{2}$ using the chain - rule is $12yy^\prime$. The derivative of the right - hand side (a constant 13) is 0. So, $12x + y+xy^\prime+12yy^\prime=0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$: $xy^\prime+12yy^\prime=-12x - y$. Factor out $y^\prime$: $y^\prime(x + 12y)=-12x - y$. Then $y^\prime=\frac{-12x - y}{x + 12y}$.

Step3: Find the slope at the point $(1,1)$

Substitute $x = 1$ and $y = 1$ into $y^\prime$: $y^\prime\big|_{(1,1)}=\frac{-12\times1 - 1}{1+12\times1}=\frac{-12 - 1}{1 + 12}=\frac{-13}{13}=-1$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m=-1$. $y - 1=-1(x - 1)$. Expand to get $y-1=-x + 1$. Then $y=-x+2$.

Answer:

$y=-x + 2$