Given the equation 

StartRoot x squared plus 6 xy plus y Superscript 6 EndRoot equals 2x2+6xy+y6=2​, 

evaluate 

StartFraction dy Over dx EndFractiondydx. 

Assume that the equation implicitly defines y as a differentiable function of x.

Question content area bottom

Part 1

If 

​F(x,y)equals=StartRoot x squared plus 6 xy plus y Superscript 6 EndRoot minus 2 equals 0x2+6xy+y6−2=0​, 

then 

Upper F Subscript xFxequals=StartFraction nothing Over nothing EndFraction .

We are tasked with finding the derivative \(\frac{dy}{dx}\) assuming that the equation defines \(y\) implicitly as a function of \(x\). Let's break this down step by step.

### Step 1: Understand the given equation

We are given the equation:

\[
\sqrt{x^2 + 6xy + y^6} = 2
\]

We can square both sides to eliminate the square root:

\[
x^2 + 6xy + y^6 = 4
\]

Now, let's define the function \(F(x, y)\) as:

\[
F(x, y) = x^2 + 6xy + y^6 - 4 = 0
\]

This equation implicitly defines \(y\) as a function of \(x\). We are tasked with finding the partial derivative of \(F(x, y)\) with respect to \(x\), which will be needed for implicit differentiation.

### Step 2: Partial derivatives of \(F(x, y)\)

We need to compute the partial derivatives \(F_x\) and \(F_y\), where \(F_x\) is the partial derivative of \(F\) with respect to \(x\), and \(F_y\) is the partial derivative with respect to \(y\).

#### Finding \(F_x\):

\[
F(x, y) = x^2 + 6xy + y^6 - 4
\]

Differentiate \(F(x, y)\) with respect to \(x\) while treating \(y\) as a function of \(x\):

\[
F_x = \frac{\partial}{\partial x} \left( x^2 + 6xy + y^6 - 4 \right)
\]

Applying the chain rule, we get:

\[
F_x = 2x + 6y
\]

#### Finding \(F_y\):

Now, let's differentiate \(F(x, y)\) with respect to \(y\):

\[
F_y = \frac{\partial}{\partial y} \left( x^2 + 6xy + y^6 - 4 \right)
\]

This gives:

\[
F_y = 6x + 6y^5
\]

### Step 3: Implicit differentiation to find \(\frac{dy}{dx}\)

Using implicit differentiation, we know that:

\[
F_x + F_y \frac{dy}{dx} = 0
\]

Substitute the values for \(F_x\) and \(F_y\):

\[
2x + 6y + (6x + 6y^5) \frac{dy}{dx} = 0
\]

Solve for \(\frac{dy}{dx}\):

\[
(6x + 6y^5) \frac{dy}{dx} = -(2x + 6y)
\]

Now, divide both sides by \(6x + 6y^5\):

\[
\frac{dy}{dx} = \frac{-(2x + 6y)}{6x + 6y^5}
\]

### Final answer:

\[
\frac{dy}{dx} = \frac{-(2x + 6y)}{6x + 6y^5}
\]

Would you like to dive deeper into any step or have more questions on this?

---

Here are five related questions to expand your understanding:

1. How do we apply implicit differentiation to more complex equations?
2. What are partial derivatives, and how are they useful in multivariable calculus?
3. How does squaring both sides of the equation help in solving the problem?
4. Why is it important to treat \(y\) as a function of \(x\) during differentiation?
5. Can we solve for \(y\) explicitly from the original equation?

**Tip**: Always check if squaring an equation introduces extraneous solutions!

Given the equation sqrt(x^2 + 6xy + y^6) = 2, evaluate dy/dx. Assume that the equation implicitly defines y as a differentiable function of x.

Learn how to apply implicit differentiation to the equation sqrt(x^2 + 6xy + y^6) = 2 to evaluate dy/dx. This step-by-step guide covers key concepts in multivariable calculus, including partial derivatives and the chain rule, making it ideal for students in Grades 11-12 and college calculus.

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