Math Problem Statement

Given x^5+5xy-y^2=y+7, use implicit differentiation to find dy/dx. dy/dx=?

Solution

We are given the equation:

x5+5xyy2=y+7x^5 + 5xy - y^2 = y + 7

To find dydx\frac{dy}{dx}, we'll use implicit differentiation, treating yy as a function of xx (i.e., y=y(x)y = y(x)).

Step 1: Differentiate both sides with respect to xx

Left-hand side:

  • The derivative of x5x^5 with respect to xx is 5x45x^4.
  • The derivative of 5xy5xy involves the product rule: ddx(5xy)=5(y+xdydx)=5y+5xdydx\frac{d}{dx}(5xy) = 5 \left( y + x \frac{dy}{dx} \right) = 5y + 5x \frac{dy}{dx}
  • The derivative of y2-y^2 using the chain rule: ddx(y2)=2ydydx\frac{d}{dx}(-y^2) = -2y \frac{dy}{dx}

So, differentiating the left-hand side, we get: 5x4+5y+5xdydx2ydydx5x^4 + 5y + 5x \frac{dy}{dx} - 2y \frac{dy}{dx}

Right-hand side:

  • The derivative of yy is dydx\frac{dy}{dx}.
  • The derivative of the constant 77 is 00.

So, differentiating the right-hand side, we get: dydx\frac{dy}{dx}

Step 2: Set up the equation

Now, equate the derivatives from both sides: 5x4+5y+5xdydx2ydydx=dydx5x^4 + 5y + 5x \frac{dy}{dx} - 2y \frac{dy}{dx} = \frac{dy}{dx}

Step 3: Solve for dydx\frac{dy}{dx}

First, group all the terms with dydx\frac{dy}{dx} on one side: 5x4+5y=dydx5xdydx+2ydydx5x^4 + 5y = \frac{dy}{dx} - 5x \frac{dy}{dx} + 2y \frac{dy}{dx}

Factor out dydx\frac{dy}{dx} on the right-hand side: 5x4+5y=dydx(15x+2y)5x^4 + 5y = \frac{dy}{dx} \left( 1 - 5x + 2y \right)

Finally, solve for dydx\frac{dy}{dx}: dydx=5x4+5y15x+2y\frac{dy}{dx} = \frac{5x^4 + 5y}{1 - 5x + 2y}

Thus, the derivative dydx\frac{dy}{dx} is:

5x4+5y15x+2y\boxed{\frac{5x^4 + 5y}{1 - 5x + 2y}}

Would you like further details or have any questions?


Here are 5 related questions for you to explore:

  1. How does implicit differentiation differ from explicit differentiation?
  2. Can you find d2ydx2\frac{d^2y}{dx^2} for the same equation?
  3. What would change if the equation involved a trigonometric function like sin(y)\sin(y)?
  4. How does the chain rule help in differentiating composite functions?
  5. What is the significance of the product rule in calculus?

Tip: When differentiating implicitly, always treat yy as a function of xx, and apply the chain rule when differentiating terms involving yy.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Product Rule
Algebra

Formulas

Implicit differentiation of x^n
Product rule: d(uv)/dx = u'v + uv'
Chain rule: d(f(y))/dx = f'(y) * dy/dx

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Grades 11-12, Calculus I