Math Problem Statement

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 .

Solution

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

Step 1: Understand the given equation

We are given the equation:

x2+6xy+y6=2\sqrt{x^2 + 6xy + y^6} = 2

We can square both sides to eliminate the square root:

x2+6xy+y6=4x^2 + 6xy + y^6 = 4

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

F(x,y)=x2+6xy+y64=0F(x, y) = x^2 + 6xy + y^6 - 4 = 0

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

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

We need to compute the partial derivatives FxF_x and FyF_y, where FxF_x is the partial derivative of FF with respect to xx, and FyF_y is the partial derivative with respect to yy.

Finding FxF_x:

F(x,y)=x2+6xy+y64F(x, y) = x^2 + 6xy + y^6 - 4

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

Fx=x(x2+6xy+y64)F_x = \frac{\partial}{\partial x} \left( x^2 + 6xy + y^6 - 4 \right)

Applying the chain rule, we get:

Fx=2x+6yF_x = 2x + 6y

Finding FyF_y:

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

Fy=y(x2+6xy+y64)F_y = \frac{\partial}{\partial y} \left( x^2 + 6xy + y^6 - 4 \right)

This gives:

Fy=6x+6y5F_y = 6x + 6y^5

Step 3: Implicit differentiation to find dydx\frac{dy}{dx}

Using implicit differentiation, we know that:

Fx+Fydydx=0F_x + F_y \frac{dy}{dx} = 0

Substitute the values for FxF_x and FyF_y:

2x+6y+(6x+6y5)dydx=02x + 6y + (6x + 6y^5) \frac{dy}{dx} = 0

Solve for dydx\frac{dy}{dx}:

(6x+6y5)dydx=(2x+6y)(6x + 6y^5) \frac{dy}{dx} = -(2x + 6y)

Now, divide both sides by 6x+6y56x + 6y^5:

dydx=(2x+6y)6x+6y5\frac{dy}{dx} = \frac{-(2x + 6y)}{6x + 6y^5}

Final answer:

dydx=(2x+6y)6x+6y5\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 yy as a function of xx during differentiation?
  5. Can we solve for yy explicitly from the original equation?

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

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Multivariable Calculus
Partial Derivatives

Formulas

sqrt(x^2 + 6xy + y^6) = 2
Implicit Differentiation: Fx + Fy(dy/dx) = 0
dy/dx = -(2x + 6y) / (6x + 6y^5)

Theorems

Chain Rule for Implicit Differentiation
Partial Derivatives

Suitable Grade Level

Grades 11-12, College Calculus