Find the equation of the tangent line to the curve y^5 = x^3 - y at the point (2, √2).

Steps to Solve

1. Differentiate the curve equation

Start with the curve equation given:

$$ y^5 = x^3 - y. $$

Differentiate both sides with respect to $x$. Using implicit differentiation, we get:

$$ 5y^4 \frac{dy}{dx} = 3x^2 - \frac{dy}{dx}. $$

2. Isolate $\frac{dy}{dx}$

Rearranging the equation to solve for $\frac{dy}{dx}$:

$$ 5y^4 \frac{dy}{dx} + \frac{dy}{dx} = 3x^2 $$

This simplifies to:

$$ (5y^4 + 1) \frac{dy}{dx} = 3x^2 $$

Hence,

$$ \frac{dy}{dx} = \frac{3x^2}{5y^4 + 1}. $$

3. Substitute the point into the equation

Now, substitute the coordinates of the point $(2, \sqrt{2})$ into the derivative to find the slope of the tangent:

$$ \frac{dy}{dx} = \frac{3(2)^2}{5(\sqrt{2})^4 + 1} $$

Calculating this gives:

$$ (2)^2 = 4; \quad (\sqrt{2})^4 = 4 $$

So,

$$ \frac{dy}{dx} = \frac{3 \cdot 4}{5 \cdot 4 + 1} = \frac{12}{20 + 1} = \frac{12}{21} = \frac{4}{7}. $$

4. Use point-slope form to find the tangent line

Now using the point-slope form of the line equation, which is:

$$ y - y_1 = m(x - x_1), $$

where $m = \frac{4}{7}$, $x_1 = 2$, and $y_1 = \sqrt{2}$:

$$ y - \sqrt{2} = \frac{4}{7}(x - 2). $$

5. Rearranging to get the equation of the tangent line

Distributing and rearranging gives:

$$ y = \frac{4}{7}x - \frac{8}{7} + \sqrt{2}. $$

This can be left in this form as the equation of the tangent line.