Which of the following points lies on the tangent to the curve $$x^4 e^y + 2\sqrt{y+1} = 3$$ at the point $$(1, 0)$$?

We start with the curve

$$x^{4}e^{y}+2\sqrt{y+1}=3.$$

We must find the equation of the tangent to this curve at the point $$(1,0)$$ and then check which of the given points satisfies that tangent equation.

Step 1 - Differentiate the curve implicitly.

The rule we use is: if $$F(x,y)=\text{constant},$$ then $$\dfrac{dF}{dx}=0,$$ and for every term containing $$y$$ we multiply by $$\dfrac{dy}{dx}.$$

Write

$$F(x,y)=x^{4}e^{y}+2\sqrt{y+1}-3.$$

Because $$F(x,y)=0,$$ we have

$$\dfrac{dF}{dx}=0.$$

Differentiate each term with respect to $$x$$:

• For $$x^{4}e^{y}$$ we use the product rule $$\dfrac{d(uv)}{dx}=u'v+uv'$$ with $$u=x^{4},\;v=e^{y}.$$ So

$$\dfrac{d}{dx}\bigl(x^{4}e^{y}\bigr)=4x^{3}e^{y}+x^{4}e^{y}\dfrac{dy}{dx}.$$

• For $$2\sqrt{y+1}=2(y+1)^{1/2}$$ we use the chain rule $$\dfrac{d}{dx}(g(y))=g'(y)\dfrac{dy}{dx}.$$ Here

$$\dfrac{d}{dx}\Bigl(2(y+1)^{1/2}\Bigr)=2\cdot\dfrac{1}{2}(y+1)^{-1/2}\dfrac{dy}{dx} =\dfrac{1}{\sqrt{y+1}}\dfrac{dy}{dx}.$$

• The derivative of the constant $$-3$$ is $$0.$$

Putting these together and equating to zero we obtain

$$4x^{3}e^{y}+x^{4}e^{y}\dfrac{dy}{dx}+\dfrac{1}{\sqrt{y+1}}\dfrac{dy}{dx}=0.$$

Step 2 - Evaluate the derivative at the given point $$(1,0).$$

Substituting $$x=1,\;y=0$$ gives:

$$4(1)^{3}e^{0}+ (1)^{4}e^{0}\dfrac{dy}{dx}+ \dfrac{1}{\sqrt{0+1}}\dfrac{dy}{dx}=0.$$

We simplify each piece:

$$4(1)^{3}e^{0}=4\cdot1\cdot1=4,$$

$$(1)^{4}e^{0}\dfrac{dy}{dx}=1\cdot1\cdot\dfrac{dy}{dx}= \dfrac{dy}{dx},$$

and

$$\dfrac{1}{\sqrt{0+1}}\dfrac{dy}{dx}= \dfrac{1}{1}\dfrac{dy}{dx}= \dfrac{dy}{dx}.$$

Therefore

$$4+\dfrac{dy}{dx}+\dfrac{dy}{dx}=0\; \Longrightarrow\; 4+2\dfrac{dy}{dx}=0.$$

Solving for $$\dfrac{dy}{dx}$$ gives

$$2\dfrac{dy}{dx}=-4\; \Longrightarrow\; \dfrac{dy}{dx}=-2.$$

Thus the slope of the tangent at $$(1,0)$$ is $$m=-2.$$

Step 3 - Write the tangent line.

Using the point-slope form $$y-y_{1}=m(x-x_{1}),$$ with $$(x_{1},y_{1})=(1,0)$$ and $$m=-2,$$ we get

$$y-0=-2(x-1).$$

Simplifying,

$$y=-2x+2.$$

Step 4 - Check each option.

We substitute the coordinates of each point into $$y=-2x+2.$$ If the equality holds, the point lies on the tangent.

A. For $$(2,2):$$ $$y=-2(2)+2=-4+2=-2\neq2.$$ Not on the line.

B. For $$(2,6):$$ $$y=-2(2)+2=-2\neq6.$$ Not on the line.

C. For $$(-2,6):$$ $$y=-2(-2)+2=4+2=6,$$ which matches $$y=6.$$ This point is on the tangent.

D. For $$(-2,4):$$ $$y=-2(-2)+2=6\neq4.$$ Not on the line.

Only option C satisfies the tangent equation.

Hence, the correct answer is Option C.