Question

find an equation of the line tangent to the curve defined by $x^{2}+2xy + y^{6}=4$ at the point $(1,1)$.

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $x^{2}+2xy + y^{6}=4$ with respect to $x$.
Using the sum - rule and product - rule:
The derivative of $x^{2}$ with respect to $x$ is $2x$.
For $2xy$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2x$ and $v = y$, we get $2y + 2x\frac{dy}{dx}$.
The derivative of $y^{6}$ with respect to $x$ is $6y^{5}\frac{dy}{dx}$, and the derivative of the constant 4 is 0.
So, $2x+2y + 2x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ together:
$2x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=-2x - 2y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x + 6y^{5})=-2x - 2y$.
Then $\frac{dy}{dx}=\frac{-2x - 2y}{2x+6y^{5}}=\frac{-x - y}{x + 3y^{5}}$.

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

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,1)}=\frac{-1 - 1}{1+3\times1^{5}}=\frac{-2}{4}=-\frac{1}{2}$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m =-\frac{1}{2}$.
$y - 1=-\frac{1}{2}(x - 1)$.
Expand to get $y-1=-\frac{1}{2}x+\frac{1}{2}$.
Then $y=-\frac{1}{2}x+\frac{1}{2}+1=-\frac{1}{2}x+\frac{3}{2}$.
Multiply through by 2 to get $2y=-x + 3$ or $x+2y=3$.

Answer:

$x + 2y=3$