Question

find an equation of the line tangent to the curve defined by $x^{5}+3xy + y^{6}=5$ at the point $(1,1)$.

Explanation:

Step1: Differentiate the equation implicitly.

Differentiate $x^{5}+3xy + y^{6}=5$ with respect to $x$.
Using the power - rule and product - rule:
The derivative of $x^{5}$ with respect to $x$ is $5x^{4}$.
For the term $3xy$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 3x$ and $v = y$, we have $3y + 3x\frac{dy}{dx}$.
The derivative of $y^{6}$ with respect to $x$ is $6y^{5}\frac{dy}{dx}$, and the derivative of the constant 5 is 0.
So, $5x^{4}+3y + 3x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ together:
$3x\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=-5x^{4}-3y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3x + 6y^{5})=-5x^{4}-3y$.
Then $\frac{dy}{dx}=\frac{-5x^{4}-3y}{3x + 6y^{5}}$.

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

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,1)}=\frac{-5(1)^{4}-3(1)}{3(1)+6(1)^{5}}=\frac{-5 - 3}{3 + 6}=\frac{-8}{9}$.

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

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m =-\frac{8}{9}$.
$y - 1=-\frac{8}{9}(x - 1)$.
Expand: $y-1=-\frac{8}{9}x+\frac{8}{9}$.
$y=-\frac{8}{9}x+\frac{8}{9}+1$.
$y=-\frac{8}{9}x+\frac{17}{9}$.

Answer:

$y =-\frac{8}{9}x+\frac{17}{9}$