find dy/dx by implicit differentiation x^4+x^2y^2+y^3=5

We are asked to find \(\frac{dy}{dx}\) for the equation:

\[
x^4 + x^2 y^2 + y^3 = 5
\]

### Step-by-step Implicit Differentiation:

1. Differentiate both sides of the equation with respect to \(x\), remembering that \(y\) is a function of \(x\) (so you'll need to use the chain rule when differentiating terms involving \(y\)).

\[
\frac{d}{dx}(x^4) + \frac{d}{dx}(x^2 y^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5)
\]

2. Differentiate each term:
   - For \(x^4\), we have:
     \[
     \frac{d}{dx}(x^4) = 4x^3
     \]

   - For \(x^2 y^2\), use the product rule. Let \(u = x^2\) and \(v = y^2\):
     \[
     \frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(u v) = u \frac{dv}{dx} + v \frac{du}{dx}
     \]
     Now differentiate:
     \[
     \frac{d}{dx}(x^2) = 2x
     \]
     and
     \[
     \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}
     \]
     So, applying the product rule:
     \[
     \frac{d}{dx}(x^2 y^2) = x^2 (2y \frac{dy}{dx}) + y^2 (2x)
     \]

   - For \(y^3\), use the chain rule:
     \[
     \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}
     \]

   - The derivative of 5 is 0, since it's a constant.

3. Substitute all the differentiated terms back into the original equation:

\[
4x^3 + \left[ x^2 (2y \frac{dy}{dx}) + y^2 (2x) \right] + 3y^2 \frac{dy}{dx} = 0
\]

4. Now, group the terms that involve \(\frac{dy}{dx}\) together:

\[
4x^3 + 2x y^2 + 2x^2 y \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0
\]

5. Factor out \(\frac{dy}{dx}\) on the left side:

\[
4x^3 + 2x y^2 + \left( 2x^2 y + 3y^2 \right) \frac{dy}{dx} = 0
\]

6. Solve for \(\frac{dy}{dx}\):

\[
\left( 2x^2 y + 3y^2 \right) \frac{dy}{dx} = - (4x^3 + 2x y^2)
\]

\[
\frac{dy}{dx} = \frac{- (4x^3 + 2x y^2)}{2x^2 y + 3y^2}
\]

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

\[
\boxed{\frac{dy}{dx} = \frac{- (4x^3 + 2x y^2)}{2x^2 y + 3y^2}}
\]

Would you like any further clarification on the steps?

### Related Questions:
1. How do you apply the product rule in implicit differentiation?
2. What is the chain rule and how is it used for implicit differentiation?
3. How do you handle higher-order derivatives in implicit differentiation?
4. What happens if the original equation includes trigonometric functions of \(y\)?
5. Can implicit differentiation be used for parametric equations?

### Tip:
Always remember to apply the chain rule carefully when differentiating terms involving both \(x\) and \(y\).

find dy/dx by implicit differentiation x^4 + x^2 y^2 + y^3 = 5

Learn how to find dy/dx using implicit differentiation for the equation x^4 + x^2y^2 + y^3 = 5. This step-by-step guide covers key calculus concepts such as the chain rule and product rule. Ideal for students in Grades 11-12 or early college-level calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation