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Find [tex]y^{\prime}[/tex] and the slope of the tangent line to the graph of the equation below at the indicated point.

[tex]\sqrt{12+y^2}-x^3+4=0 \quad (2,2)[/tex]

[tex]\[
\begin{array}{l}
y^{\prime}=\square \\
\left.y^{\prime}\right|_{(2,2)}=\square \text{ (Simplify your answer.) }
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To solve for [tex]\( y' \)[/tex] and find the slope of the tangent line at the point [tex]\((2, 2)\)[/tex] for the given equation

[tex]\[ \sqrt{12 + y^2} - x^3 + 4 = 0, \][/tex]

follow these steps:

### Step 1: Implicit Differentiation
Differentiate both sides of the equation with respect to [tex]\( x \)[/tex]:

[tex]\[ \frac{d}{dx} \left( \sqrt{12 + y^2} - x^3 + 4 \right) = \frac{d}{dx}(0). \][/tex]

The left-hand side simplifies using the chain rule.

1. For the term [tex]\( \sqrt{12 + y^2} \)[/tex]:

[tex]\[ \frac{d}{dx} \left( \sqrt{12 + y^2} \right) = \frac{1}{2\sqrt{12 + y^2}} \cdot \frac{d}{dx}(12 + y^2) = \frac{1}{2\sqrt{12 + y^2}} \cdot 2y \cdot \frac{dy}{dx}. \][/tex]

Therefore,

[tex]\[ \frac{d}{dx} \left( \sqrt{12 + y^2} \right) = \frac{y}{\sqrt{12 + y^2}} \cdot y'. \][/tex]

2. For the term [tex]\( -x^3 \)[/tex]:

[tex]\[ \frac{d}{dx} \left( -x^3 \right) = -3x^2. \][/tex]

3. The term [tex]\( +4 \)[/tex] is a constant, so its derivative is [tex]\( 0 \)[/tex].

Putting it all together, we have:

[tex]\[ \frac{y}{\sqrt{12 + y^2}} \cdot y' - 3x^2 = 0. \][/tex]

### Step 2: Solve for [tex]\( y' \)[/tex]
To isolate [tex]\( y' \)[/tex]:

[tex]\[ \frac{y}{\sqrt{12 + y^2}} \cdot y' = 3x^2. \][/tex]

[tex]\[ y' = \frac{3x^2 \sqrt{12 + y^2}}{y}. \][/tex]

### Step 3: Find [tex]\( y' \)[/tex] at the Point [tex]\( (2, 2) \)[/tex]
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 2 \)[/tex] into the equation for [tex]\( y' \)[/tex]:

[tex]\[ y' = \frac{3(2)^2 \sqrt{12 + (2)^2}}{2}. \][/tex]

Calculate each part:

[tex]\[ (2)^2 = 4, \][/tex]
[tex]\[ 12 + (2)^2 = 12 + 4 = 16, \][/tex]
[tex]\[ \sqrt{16} = 4. \][/tex]

Now substitute back:

[tex]\[ y' = \frac{3 \cdot 4 \cdot 4}{2} = \frac{48}{2} = 24. \][/tex]

### Summary
Therefore, the derivative [tex]\( y' \)[/tex] in general form is:

[tex]\[ y' = \frac{3x^2 \sqrt{12 + y^2}}{y}, \][/tex]

and at the point [tex]\( (2, 2) \)[/tex], the value of [tex]\( y' \)[/tex] (the slope of the tangent line) is:

[tex]\[ \left.y^{\prime}\right|_{(2, 2)} = 24. \][/tex]

So,

[tex]\[ \begin{array}{l} y^{\prime}=\frac{3x^2 \sqrt{12 + y^2}}{y} \\ \left.y^{\prime}\right|_{(2,2)}=24 \end{array} \][/tex]
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