Math Problem Statement

Suppose that x and y are related by the given equation and use implicit differentiation to determine dy/dx: x^5*y + y^5*x = 7.

Solution

To solve this problem using implicit differentiation, we aim to find dydx\frac{dy}{dx}. Given the equation:

x5y+y5x=7,x^5y + y^5x = 7,

we will differentiate both sides of the equation with respect to xx, keeping in mind that yy is a function of xx.

Step-by-step Solution:

  1. Differentiate x5yx^5y using the product rule: ddx(x5y)=ddx(x5)y+x5ddx(y),\frac{d}{dx}(x^5y) = \frac{d}{dx}(x^5) \cdot y + x^5 \cdot \frac{d}{dx}(y), which becomes: 5x4y+x5dydx.5x^4y + x^5\frac{dy}{dx}.

  2. Differentiate y5xy^5x also using the product rule: ddx(y5x)=ddx(y5)x+y5ddx(x),\frac{d}{dx}(y^5x) = \frac{d}{dx}(y^5) \cdot x + y^5 \cdot \frac{d}{dx}(x), which becomes: 5y4dydxx+y5.5y^4\frac{dy}{dx} \cdot x + y^5.

  3. Differentiate the right-hand side, 77, which is constant: ddx(7)=0.\frac{d}{dx}(7) = 0.

Now, combining everything, we have: 5x4y+x5dydx+5y4xdydx+y5=0.5x^4y + x^5\frac{dy}{dx} + 5y^4x\frac{dy}{dx} + y^5 = 0.

  1. Group the terms involving dydx\frac{dy}{dx} on one side: x5dydx+5y4xdydx=5x4yy5.x^5\frac{dy}{dx} + 5y^4x\frac{dy}{dx} = -5x^4y - y^5.

Factor out dydx\frac{dy}{dx}: dydx(x5+5y4x)=5x4yy5.\frac{dy}{dx}(x^5 + 5y^4x) = -5x^4y - y^5.

  1. Solve for dydx\frac{dy}{dx}: dydx=5x4yy5x5+5y4x.\frac{dy}{dx} = \frac{-5x^4y - y^5}{x^5 + 5y^4x}.

Final Answer:

dydx=5x4yy5x5+5y4x.\frac{dy}{dx} = \frac{-5x^4y - y^5}{x^5 + 5y^4x}.

Let me know if you'd like further clarification or need help applying this to specific values.


Related Questions:

  1. What is the significance of implicit differentiation in finding dydx\frac{dy}{dx}?
  2. How does the product rule simplify differentiation in equations involving two functions?
  3. What happens to the equation if x=0x = 0 or y=0y = 0?
  4. How can we confirm the correctness of this differentiation result?
  5. What applications of implicit differentiation exist in physics or engineering?

Tip:

When differentiating implicitly, always remember to apply the chain rule when differentiating yy, since yy is a function of xx.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Algebra

Formulas

Product rule: (uv)' = u'v + uv'
Chain rule for implicit differentiation: d/dx [f(y)] = f'(y) * dy/dx

Theorems

Implicit Differentiation Theorem
Product Rule

Suitable Grade Level

Grades 11-12 (Calculus)