Math Problem Statement

Differentiate the implicit of xy^2+3y^3=5cos(4x)+2e^2x + In y

Solution

To differentiate the implicit equation xy2+3y3=5cos(4x)+2e2x+lnyxy^2 + 3y^3 = 5\cos(4x) + 2e^{2x} + \ln y with respect to xx, we’ll apply implicit differentiation, treating yy as a function of xx (i.e., y=y(x)y = y(x)).

  1. Differentiate each term on both sides with respect to xx:

    ddx(xy2+3y3)=ddx(5cos(4x)+2e2x+lny)\frac{d}{dx} \left( xy^2 + 3y^3 \right) = \frac{d}{dx} \left( 5\cos(4x) + 2e^{2x} + \ln y \right)

  2. Differentiate each term on the left side:

    • For xy2xy^2: Use the product rule: ddx(xy2)=xddx(y2)+y2ddx(x)=x(2yy)+y2=2xyy+y2\frac{d}{dx}(xy^2) = x \cdot \frac{d}{dx}(y^2) + y^2 \cdot \frac{d}{dx}(x) = x \cdot (2y \cdot y') + y^2 = 2xy y' + y^2

    • For 3y33y^3: ddx(3y3)=33y2y=9y2y\frac{d}{dx}(3y^3) = 3 \cdot 3y^2 \cdot y' = 9y^2 y'

    So, the derivative of the left side is: 2xyy+y2+9y2y2xy y' + y^2 + 9y^2 y'

  3. Differentiate each term on the right side:

    • For 5cos(4x)5\cos(4x): ddx(5cos(4x))=5(sin(4x))4=20sin(4x)\frac{d}{dx}(5\cos(4x)) = 5 \cdot (-\sin(4x)) \cdot 4 = -20\sin(4x)

    • For 2e2x2e^{2x}: ddx(2e2x)=2e2x2=4e2x\frac{d}{dx}(2e^{2x}) = 2 \cdot e^{2x} \cdot 2 = 4e^{2x}

    • For lny\ln y: ddx(lny)=1yy=yy\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot y' = \frac{y'}{y}

    Thus, the derivative of the right side is: 20sin(4x)+4e2x+yy-20\sin(4x) + 4e^{2x} + \frac{y'}{y}

  4. Combine terms:

    We now have: 2xyy+y2+9y2y=20sin(4x)+4e2x+yy2xy y' + y^2 + 9y^2 y' = -20\sin(4x) + 4e^{2x} + \frac{y'}{y}

  5. Solve for yy':

    Collect all yy' terms on one side: y(2xy+9y21y)=20sin(4x)+4e2xy2y' (2xy + 9y^2 - \frac{1}{y}) = -20\sin(4x) + 4e^{2x} - y^2

    Finally, solving for yy': y=20sin(4x)+4e2xy22xy+9y21yy' = \frac{-20\sin(4x) + 4e^{2x} - y^2}{2xy + 9y^2 - \frac{1}{y}}

The derivative yy' is: y=20sin(4x)+4e2xy22xy+9y21yy' = \frac{-20\sin(4x) + 4e^{2x} - y^2}{2xy + 9y^2 - \frac{1}{y}}

Would you like further clarification on any part?


Here are five related questions you might find helpful:

  1. How would the differentiation change if yy were considered independent of xx?
  2. What are the product and chain rules for derivatives?
  3. How do you differentiate trigonometric functions?
  4. How do you solve implicit differentiation problems with natural logarithms?
  5. What are the key steps for finding derivatives involving exponential functions?

Tip: Always apply the chain rule carefully when differentiating functions that include yy as a function of xx.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Differentiation of Exponential Functions
Differentiation of Trigonometric Functions
Differentiation of Logarithmic Functions

Formulas

Product Rule: (uv)' = u'v + uv'
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Derivative of cos(x): d/dx[cos(x)] = -sin(x)
Derivative of e^x: d/dx[e^x] = e^x
Derivative of ln(x): d/dx[ln(x)] = 1/x

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12