Math Problem Statement

use implicit differentiation to find dy/dx of 5x^3=-3xy+2 in terms of y

Solution

To find dydx\frac{dy}{dx} using implicit differentiation for the equation:

5x3=3xy+25x^3 = -3xy + 2

Step 1: Differentiate both sides with respect to xx.

  1. Differentiate the left side: ddx(5x3)=15x2\frac{d}{dx}(5x^3) = 15x^2

  2. Differentiate the right side: Apply the product rule to the term 3xy-3xy and differentiate 22, which is constant:

    • The derivative of 3xy-3xy using the product rule is: ddx(3xy)=3(xddx(y)+yddx(x))=3(xdydx+y)\frac{d}{dx}(-3xy) = -3 \left( x \frac{d}{dx}(y) + y \frac{d}{dx}(x) \right) = -3(x \frac{dy}{dx} + y)
    • The derivative of the constant 22 is 00.

So, the differentiated equation becomes: 15x2=3(xdydx+y)15x^2 = -3(x \frac{dy}{dx} + y)

Step 2: Solve for dydx\frac{dy}{dx}.

Now, isolate dydx\frac{dy}{dx}: 15x2=3xdydx3y15x^2 = -3x \frac{dy}{dx} - 3y Add 3y3y to both sides: 15x2+3y=3xdydx15x^2 + 3y = -3x \frac{dy}{dx} Now, divide both sides by 3x-3x: dydx=15x23y3x\frac{dy}{dx} = \frac{-15x^2 - 3y}{-3x}

Simplify: dydx=15x2+3y3x\frac{dy}{dx} = \frac{15x^2 + 3y}{3x}

Finally, factor out the common factor of 3: dydx=5x2+yx\frac{dy}{dx} = \frac{5x^2 + y}{x}

Final Answer:

dydx=5x2+yx\frac{dy}{dx} = \frac{5x^2 + y}{x}

Would you like more details on any part of this process? Here's a list of related questions to further explore:

  1. What are the product and chain rules used for implicit differentiation?
  2. How do you apply implicit differentiation to more complex equations?
  3. What happens if the equation includes higher-order terms like x4x^4 or trigonometric functions?
  4. How do you check if an implicit derivative is correct after differentiating?
  5. What is the geometric interpretation of implicit differentiation in terms of slopes?

Tip: Remember that in implicit differentiation, when differentiating terms involving yy, treat yy as a function of xx, meaning you apply the chain rule (ddxy=dydx\frac{d}{dx}y = \frac{dy}{dx}).

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Algebraic Manipulation

Formulas

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

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12