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Answer :
To solve this problem, we need to find the slope of the tangent line to the curve described by the equation [tex]\(2x^4 - y^2 = 23\)[/tex] at the point [tex]\((2, 3)\)[/tex]. We use implicit differentiation to find the derivative [tex]\(\frac{dy}{dx}\)[/tex], which represents the slope of the tangent line.
### Step-by-step Solution:
1. Differentiate the Equation Implicitly:
Start with the given equation:
[tex]\[
2x^4 - y^2 = 23
\][/tex]
Differentiate both sides with respect to [tex]\(x\)[/tex]. Remember, when differentiating [tex]\(y^2\)[/tex], treat [tex]\(y\)[/tex] as a function of [tex]\(x\)[/tex] and use the chain rule:
[tex]\[
\frac{d}{dx}(2x^4) - \frac{d}{dx}(y^2) = \frac{d}{dx}(23)
\][/tex]
Differentiating each term:
- [tex]\(\frac{d}{dx}(2x^4) = 8x^3\)[/tex]
- [tex]\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)[/tex] (using the chain rule)
- [tex]\(\frac{d}{dx}(23) = 0\)[/tex]
Plug these into the differentiated equation:
[tex]\[
8x^3 - 2y \cdot \frac{dy}{dx} = 0
\][/tex]
2. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
Rearrange the equation to solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
8x^3 = 2y \cdot \frac{dy}{dx}
\][/tex]
Divide both sides by [tex]\(2y\)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{8x^3}{2y} = \frac{4x^3}{y}
\][/tex]
3. Evaluate [tex]\(\frac{dy}{dx}\)[/tex] at the Point [tex]\((2, 3)\)[/tex]:
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 3\)[/tex] into the derivative:
[tex]\[
\frac{dy}{dx} \Big|_{(2,3)} = \frac{4(2)^3}{3} = \frac{4 \times 8}{3} = \frac{32}{3}
\][/tex]
Therefore, the slope of the tangent line at the point [tex]\((2, 3)\)[/tex] is [tex]\(\frac{32}{3}\)[/tex].
### Step-by-step Solution:
1. Differentiate the Equation Implicitly:
Start with the given equation:
[tex]\[
2x^4 - y^2 = 23
\][/tex]
Differentiate both sides with respect to [tex]\(x\)[/tex]. Remember, when differentiating [tex]\(y^2\)[/tex], treat [tex]\(y\)[/tex] as a function of [tex]\(x\)[/tex] and use the chain rule:
[tex]\[
\frac{d}{dx}(2x^4) - \frac{d}{dx}(y^2) = \frac{d}{dx}(23)
\][/tex]
Differentiating each term:
- [tex]\(\frac{d}{dx}(2x^4) = 8x^3\)[/tex]
- [tex]\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)[/tex] (using the chain rule)
- [tex]\(\frac{d}{dx}(23) = 0\)[/tex]
Plug these into the differentiated equation:
[tex]\[
8x^3 - 2y \cdot \frac{dy}{dx} = 0
\][/tex]
2. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
Rearrange the equation to solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
8x^3 = 2y \cdot \frac{dy}{dx}
\][/tex]
Divide both sides by [tex]\(2y\)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{8x^3}{2y} = \frac{4x^3}{y}
\][/tex]
3. Evaluate [tex]\(\frac{dy}{dx}\)[/tex] at the Point [tex]\((2, 3)\)[/tex]:
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 3\)[/tex] into the derivative:
[tex]\[
\frac{dy}{dx} \Big|_{(2,3)} = \frac{4(2)^3}{3} = \frac{4 \times 8}{3} = \frac{32}{3}
\][/tex]
Therefore, the slope of the tangent line at the point [tex]\((2, 3)\)[/tex] is [tex]\(\frac{32}{3}\)[/tex].
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