https://file.aiquickdraw.com/mathbot/file/1730681315253to5gxmk3.jpg

To find \(\frac{dy}{dx}\) for the equation given:

\[
x^3 \cdot y + 5 \cdot x \cdot y^5 = 3
\]

we'll use **implicit differentiation** with respect to \( x \).

### Step-by-Step Solution

1. **Differentiate both sides with respect to \( x \):**
   \[
   \frac{d}{dx}(x^3 \cdot y + 5 \cdot x \cdot y^5) = \frac{d}{dx}(3)
   \]

2. **Apply the product rule** on each term of the left side. For the first term \( x^3 \cdot y \):
   \[
   \frac{d}{dx}(x^3 \cdot y) = \frac{d}{dx}(x^3) \cdot y + x^3 \cdot \frac{dy}{dx} = 3x^2 \cdot y + x^3 \cdot \frac{dy}{dx}
   \]

3. **Apply the product rule** on the second term \( 5 \cdot x \cdot y^5 \):
   \[
   \frac{d}{dx}(5 \cdot x \cdot y^5) = 5 \cdot \frac{d}{dx}(x) \cdot y^5 + 5 \cdot x \cdot \frac{d}{dx}(y^5)
   \]
   - Differentiate \( x \) to get \( 1 \), so the first part is \( 5 \cdot y^5 \).
   - Differentiate \( y^5 \) to get \( 5y^4 \cdot \frac{dy}{dx} \), so the second part is \( 5 \cdot x \cdot 5y^4 \cdot \frac{dy}{dx} = 25xy^4 \cdot \frac{dy}{dx} \).

   Putting it together:
   \[
   \frac{d}{dx}(5 \cdot x \cdot y^5) = 5y^5 + 25xy^4 \cdot \frac{dy}{dx}
   \]

4. **Substitute back into the equation:**
   \[
   3x^2 \cdot y + x^3 \cdot \frac{dy}{dx} + 5y^5 + 25xy^4 \cdot \frac{dy}{dx} = 0
   \]

5. **Isolate \( \frac{dy}{dx} \):**
   Combine all terms involving \( \frac{dy}{dx} \):
   \[
   x^3 \cdot \frac{dy}{dx} + 25xy^4 \cdot \frac{dy}{dx} = -3x^2 \cdot y - 5y^5
   \]

6. **Factor out \( \frac{dy}{dx} \):**
   \[
   \frac{dy}{dx} (x^3 + 25xy^4) = -3x^2 \cdot y - 5y^5
   \]

7. **Solve for \( \frac{dy}{dx} \):**
   \[
   \frac{dy}{dx} = \frac{-3x^2 \cdot y - 5y^5}{x^3 + 25xy^4}
   \]

### Final Answer
\[
\frac{dy}{dx} = \frac{-3x^2 \cdot y - 5y^5}{x^3 + 25xy^4}
\]

Would you like further details or have any questions on this? 

Here are some related questions:
1. How does implicit differentiation work in finding derivatives for non-explicit functions?
2. What are some practical applications of implicit differentiation?
3. How does the product rule apply to functions of two variables like \( x \) and \( y \)?
4. What would happen if we wanted to find the second derivative, \( \frac{d^2y}{dx^2} \)?
5. Why is it necessary to rearrange the equation to isolate \( \frac{dy}{dx} \) in implicit differentiation?

**Tip:** When using implicit differentiation, remember to apply the product and chain rules carefully whenever \( y \) is multiplied by functions of \( x \).

Suppose x^3 * y + 5 * x * y^5 = 3. Find dy/dx in terms of x and y.

Learn how to find dy/dx in terms of x and y using implicit differentiation for the equation x^3 * y + 5 * x * y^5 = 3. This solution covers applying the product rule and isolating dy/dx.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation