Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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]
[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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
