Math Problem Statement

Calculate the derivative y'(x) of the implicitly defined function given by 3x + 2y^3 + x^2 e^y = 4.

Solution

The problem appears to involve finding the derivative y(x)y'(x) of an implicitly defined function. Based on the text in the image, it can be translated as:


Find the derivative y(x)y'(x) of the implicitly defined function given by:

3x+2y3+x2ey=4.3x + 2y^3 + x^2 e^y = 4.


Let me solve this for you step-by-step.

Step 1: Differentiate both sides with respect to xx

Using implicit differentiation: ddx(3x+2y3+x2ey)=ddx(4).\frac{d}{dx}\big(3x + 2y^3 + x^2e^y\big) = \frac{d}{dx}(4).

On the left:

  1. For 3x3x: ddx(3x)=3\frac{d}{dx}(3x) = 3,
  2. For 2y32y^3: Apply the chain rule: ddx(2y3)=6y2dydx=6y2y\frac{d}{dx}(2y^3) = 6y^2 \frac{dy}{dx} = 6y^2 y',
  3. For x2eyx^2 e^y: Use the product rule: ddx(x2ey)=ddx(x2)ey+x2ddx(ey)=2xey+x2eydydx.\frac{d}{dx}(x^2 e^y) = \frac{d}{dx}(x^2) \cdot e^y + x^2 \cdot \frac{d}{dx}(e^y) = 2x e^y + x^2 e^y \frac{dy}{dx}.

On the right: ddx(4)=0.\frac{d}{dx}(4) = 0.

Thus, the differentiated equation becomes: 3+6y2y+2xey+x2eyy=0.3 + 6y^2 y' + 2x e^y + x^2 e^y y' = 0.


Step 2: Solve for yy'

Group the terms involving yy': 6y2y+x2eyy=(3+2xey).6y^2 y' + x^2 e^y y' = - (3 + 2x e^y).

Factor out yy' on the left: y(6y2+x2ey)=(3+2xey).y'(6y^2 + x^2 e^y) = - (3 + 2x e^y).

Solve for yy': y=(3+2xey)6y2+x2ey.y' = \frac{- (3 + 2x e^y)}{6y^2 + x^2 e^y}.


Final Answer:

The derivative is: y=(3+2xey)6y2+x2ey.y' = \frac{- (3 + 2x e^y)}{6y^2 + x^2 e^y}.

Would you like me to clarify any step or explore this further?


Related Questions:

  1. What is the geometric meaning of y(x)y'(x) in this context?
  2. How do you interpret the role of implicit differentiation in solving such problems?
  3. What happens when 6y2+x2ey=06y^2 + x^2 e^y = 0? Can yy' still be defined?
  4. Can this equation be solved explicitly for yy?
  5. How does the exponential term eye^y impact the derivative?

Tip:

Always isolate yy' systematically by grouping terms when performing implicit differentiation.

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Math Problem Analysis

Mathematical Concepts

Calculus
Implicit Differentiation
Derivatives

Formulas

Product Rule: (uv)' = u'v + uv'
Chain Rule: (f(g(x)))' = f'(g(x))g'(x)

Theorems

Derivative of Exponential Functions
Implicit Differentiation Theorem

Suitable Grade Level

Grades 11-12