Key Concepts and Formulas

• Equation of a Tangent Line: The equation of the tangent line to a curve at a point $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.
• Implicit Differentiation: A technique used to find the derivative of a function defined implicitly.
• Derivative of $e^u$: $\frac{d}{dx}(e^u) = e^u \frac{du}{dx}$
• Derivative of $\sqrt{u}$: $\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx}$

Step-by-Step Solution

Step 1: Differentiate the given equation implicitly with respect to x. We are given the equation $x^4 e^y + 2\sqrt{y+1} = 3$. We need to find $\frac{dy}{dx}$ using implicit differentiation. Differentiating both sides with respect to $x$, we get: $\frac{d}{dx}(x^4 e^y + 2\sqrt{y+1}) = \frac{d}{dx}(3)$ $\frac{d}{dx}(x^4 e^y) + \frac{d}{dx}(2\sqrt{y+1}) = 0$ Using the product rule on the first term and the chain rule on both terms: $(4x^3 e^y + x^4 e^y \frac{dy}{dx}) + 2 \cdot \frac{1}{2\sqrt{y+1}} \cdot \frac{dy}{dx} = 0$ $4x^3 e^y + x^4 e^y \frac{dy}{dx} + \frac{1}{\sqrt{y+1}} \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$. We want to isolate $\frac{dy}{dx}$. Rearrange the equation to group terms with $\frac{dy}{dx}$: $\frac{dy}{dx} (x^4 e^y + \frac{1}{\sqrt{y+1}}) = -4x^3 e^y$ $\frac{dy}{dx} = \frac{-4x^3 e^y}{x^4 e^y + \frac{1}{\sqrt{y+1}}}$ $\frac{dy}{dx} = \frac{-4x^3 e^y}{\frac{x^4 e^y \sqrt{y+1} + 1}{\sqrt{y+1}}}$ $\frac{dy}{dx} = \frac{-4x^3 e^y \sqrt{y+1}}{x^4 e^y \sqrt{y+1} + 1}$

Step 3: Evaluate $\frac{dy}{dx}$ at the point (1, 0). We are given the point (1, 0), so we substitute $x = 1$ and $y = 0$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx}\Big|_{(1, 0)} = \frac{-4(1)^3 e^0 \sqrt{0+1}}{(1)^4 e^0 \sqrt{0+1} + 1}$ $\frac{dy}{dx}\Big|_{(1, 0)} = \frac{-4(1)(1)(1)}{(1)(1)(1) + 1}$ $\frac{dy}{dx}\Big|_{(1, 0)} = \frac{-4}{1 + 1} = \frac{-4}{2} = -2$ Thus, the slope of the tangent at (1, 0) is -2.

Step 4: Find the equation of the tangent line. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (1, 0)$ and $m = -2$, we get: $y - 0 = -2(x - 1)$ $y = -2x + 2$

Step 5: Check which of the given points lies on the tangent line. We need to check which of the given points satisfy the equation $y = -2x + 2$.

• (A) (2, 2): $2 = -2(2) + 2 \Rightarrow 2 = -4 + 2 \Rightarrow 2 = -2$. This is false.

Recalculate the slope and tangent line equation: The derivative is $\frac{dy}{dx} = \frac{-4x^3 e^y \sqrt{y+1}}{x^4 e^y \sqrt{y+1} + 1}$. At (1, 0), $\frac{dy}{dx} = \frac{-4(1)^3 e^0 \sqrt{0+1}}{(1)^4 e^0 \sqrt{0+1} + 1} = \frac{-4}{1+1} = -2$. The tangent line is $y - 0 = -2(x - 1)$, which simplifies to $y = -2x + 2$.

Now check the points again:

• (A) (2, 2): $2 = -2(2) + 2 \Rightarrow 2 = -4 + 2 \Rightarrow 2 = -2$. This is FALSE.
• (B) (-2, 4): $4 = -2(-2) + 2 \Rightarrow 4 = 4 + 2 \Rightarrow 4 = 6$. This is FALSE.
• (C) (2, 6): $6 = -2(2) + 2 \Rightarrow 6 = -4 + 2 \Rightarrow 6 = -2$. This is FALSE.
• (D) (-2, 6): $6 = -2(-2) + 2 \Rightarrow 6 = 4 + 2 \Rightarrow 6 = 6$. This is TRUE.

Therefore, the point (-2, 6) lies on the tangent line.

Step 6: Re-evaluate and correct the solution

There must be an error. Let's start from differentiating the equation. Given $x^4 e^y + 2\sqrt{y+1} = 3$. Differentiating with respect to $x$, $4x^3 e^y + x^4 e^y \frac{dy}{dx} + \frac{2}{2\sqrt{y+1}} \frac{dy}{dx} = 0$ $4x^3 e^y + x^4 e^y \frac{dy}{dx} + \frac{1}{\sqrt{y+1}} \frac{dy}{dx} = 0$ $\frac{dy}{dx}(x^4 e^y + \frac{1}{\sqrt{y+1}}) = -4x^3 e^y$ $\frac{dy}{dx} = \frac{-4x^3 e^y}{x^4 e^y + \frac{1}{\sqrt{y+1}}}$ At (1, 0): $\frac{dy}{dx} = \frac{-4(1)^3 e^0}{(1)^4 e^0 + \frac{1}{\sqrt{0+1}}} = \frac{-4}{1 + 1} = -2$ The tangent line equation is $y - 0 = -2(x - 1)$, so $y = -2x + 2$.

Now check the points. (A) (2, 2): $2 = -2(2) + 2 = -2$. Incorrect. (B) (-2, 4): $4 = -2(-2) + 2 = 6$. Incorrect. (C) (2, 6): $6 = -2(2) + 2 = -2$. Incorrect. (D) (-2, 6): $6 = -2(-2) + 2 = 6$. Correct.

There was an error in the provided "Correct Answer".

The correct point is (-2, 6).

Common Mistakes & Tips

• Be careful with the chain rule when differentiating terms involving $y$. Remember to multiply by $\frac{dy}{dx}$.
• Double-check your algebraic manipulations when solving for $\frac{dy}{dx}$.
• After finding the tangent line equation, carefully substitute the coordinates of each point to see if it satisfies the equation.

Summary

We found the equation of the tangent line to the curve $x^4 e^y + 2\sqrt{y+1} = 3$ at the point (1, 0) using implicit differentiation. We calculated $\frac{dy}{dx}$ at (1, 0) to be -2, and thus the tangent line is $y = -2x + 2$. By substituting the coordinates of the given points into the equation of the tangent line, we found that the point (-2, 6) lies on the tangent line.

Final Answer

The final answer is \boxed{(-2, 6)}, which corresponds to option (D).