Find 

StartFraction dy Over dx EndFractiondydx 

at the given point.

3 x squared minus 4 y Superscript 4 Baseline equals 2 x minus 4 y3x2−4y4=2x−4y 

at left parenthesis 0 comma 1 right parenthesis(0,1)

Question content area bottom

Part 1

The value of 

StartFraction dy Over dx EndFractiondydx 

at 

left parenthesis 0 comma 1 right parenthesis(0,1) 

is 

enter your response here.

​(Simplify your​ answer.)

To find the derivative \(\frac{dy}{dx}\) of the implicitly defined function \(3x^2 - 4y^4 = 2x - 4y\) at the point \((0, 1)\), we use implicit differentiation.

Let's differentiate both sides of the given equation with respect to \(x\), applying the chain rule where necessary.

Given:
\[ 3x^2 - 4y^4 = 2x - 4y \]

We'll differentiate each term with respect to \(x\).It seems there was an issue while calculating \(\frac{dy}{dx}\). Let's go through the implicit differentiation process manually.

1. Differentiate both sides with respect to \(x\):
   - For the left-hand side \(3x^2 - 4y^4\), we get:
     \[ \frac{d}{dx}(3x^2) - \frac{d}{dx}(4y^4) = 6x - 16y^3\frac{dy}{dx} \]
   - For the right-hand side \(2x - 4y\), we get:
     \[ \frac{d}{dx}(2x) - \frac{d}{dx}(4y) = 2 - 4\frac{dy}{dx} \]

2. Set up the equation for the derivatives:
   \[ 6x - 16y^3\frac{dy}{dx} = 2 - 4\frac{dy}{dx} \]

3. Solve for \(\frac{dy}{dx}\):
   \[ 6x - 2 = -16y^3\frac{dy}{dx} + 4\frac{dy}{dx} \]
   \[ 6x - 2 = (-16y^3 + 4)\frac{dy}{dx} \]
   \[ \frac{dy}{dx} = \frac{6x - 2}{-16y^3 + 4} \]

Now, we'll evaluate \(\frac{dy}{dx}\) at the point \((0,1)\).

\[ \frac{dy}{dx} = \frac{6(0) - 2}{-16(1)^3 + 4} = \frac{-2}{-16 + 4} = \frac{-2}{-12} = \frac{1}{6} \]

Thus, the value of \(\frac{dy}{dx}\) at \((0, 1)\) is \(\frac{1}{6}\).

Would you like further explanations or have any other questions?

---

### Relative Questions:
1. What is implicit differentiation and when is it used?
2. Why do we need to apply the chain rule when differentiating implicitly?
3. How do we handle terms involving \(y\) when differentiating implicitly?
4. What is the geometric interpretation of \(\frac{dy}{dx}\) at a point on a curve?
5. How do you interpret the result of \(\frac{dy}{dx} = \frac{1}{6}\) in this context?
6. Can implicit functions always be solved for \(y\) in terms of \(x\)?
7. What happens if the denominator when solving for \(\frac{dy}{dx}\) is zero?
8. How can we check if a critical point is a maximum, minimum, or saddle point?

### Tip:
When using implicit differentiation, always remember to differentiate each term with respect to the independent variable and apply the chain rule for the dependent variable.

Learn how to find the derivative dy/dx at the point (0, 1) for the equation 3x^2 - 4y^4 = 2x - 4y using implicit differentiation. This problem involves applying the chain rule and solving for dy/dx. Suitable for college-level mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation