To find the equation of the tangent line to the curve given by $\frac{x}{y^3} + 2y^3 = 1$ at the point $(-1,1)$, we need to follow these steps:

1. Implicit Differentiation: Differentiate the given equation implicitly with respect to $x$.

2. Solve for $\frac{dy}{dx}$: Isolate $\frac{dy}{dx}$ to find the slope of the tangent line.

3. Equation of the Tangent Line: Use the point-slope form of the line equation to write the tangent line's equation.

Step-by-Step Solution:

1. Implicit Differentiation: Given the curve: $\frac{x}{y^3} + 2y^3 = 1$ Differentiate both sides with respect to $x$: $\frac{d}{dx} \left( \frac{x}{y^3} \right) + \frac{d}{dx} (2y^3) = \frac{d}{dx} (1)$

   Applying the quotient rule to $\frac{x}{y^3}$: $\frac{d}{dx} \left( \frac{x}{y^3} \right) = \frac{y^3 \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(y^3)}{(y^3)^2}$ $= \frac{y^3 \cdot 1 - x \cdot 3y^2 \frac{dy}{dx}}{y^6}$ $= \frac{y^3 - 3xy^2 \frac{dy}{dx}}{y^6}$ $= \frac{1}{y^3} - \frac{3x}{y^4} \frac{dy}{dx}$

   Differentiating $2y^3$: $\frac{d}{dx} (2y^3) = 2 \cdot 3y^2 \frac{dy}{dx} = 6y^2 \frac{dy}{dx}$

   So, the differentiated equation is: $\frac{1}{y^3} - \frac{3x}{y^4} \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = 0$

2. Solve for $\frac{dy}{dx}$: Combine terms involving $\frac{dy}{dx}$: $\frac{1}{y^3} + \left(6y^2 - \frac{3x}{y^4}\right) \frac{dy}{dx} = 0$ $\left(6y^2 - \frac{3x}{y^4}\right) \frac{dy}{dx} = -\frac{1}{y^3}$ $\frac{dy}{dx} = -\frac{\frac{1}{y^3}}{6y^2 - \frac{3x}{y^4}}$ Simplify the denominator: $\frac{dy}{dx} = -\frac{1}{y^3} \cdot \frac{y^4}{6y^6 - 3x}$ $= -\frac{y}{6y^6 - 3x}$

   Substitute $y = 1$ and $x = -1$ into the derivative: $\left.\frac{dy}{dx}\right|_{(-1, 1)} = -\frac{1}{6(1)^6 - 3(-1)}$ $= -\frac{1}{6 + 3} = -\frac{1}{9}$

3. Equation of the Tangent Line: The slope of the tangent line at $(-1, 1)$ is $-\frac{1}{9}$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 1 = -\frac{1}{9}(x + 1)$ $y - 1 = -\frac{1}{9}x - \frac{1}{9}$ $y = -\frac{1}{9}x + \frac{8}{9}$

So, the equation of the tangent line at the point $(-1, 1)$ is: $y = -\frac{1}{9}x + \frac{8}{9}$

Would you like more details or have any questions?

Related Questions:

1. How do you apply implicit differentiation to more complex functions involving trigonometric terms?
2. What is the geometric interpretation of the derivative at a point on a curve?
3. How can you find the normal line to a curve at a given point?
4. What are the steps to verify if a point lies on a given curve?
5. How do you compute higher-order derivatives for implicitly defined functions?

Tip:

When using implicit differentiation, always remember to treat $y$ as a function of $x$ and apply the chain rule accordingly.