implicit differentiation: x^2y^3 +4xy + x - 6y =2

To perform implicit differentiation on the equation:

\[
x^2y^3 + 4xy + x - 6y = 2
\]

we need to differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., \( y = y(x) \)). We'll apply the product rule and chain rule where necessary. Let's differentiate each term:

### Step 1: Differentiate the first term \( x^2y^3 \)
We use the product rule, which states:
\[
\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
\]
So, for \( x^2y^3 \):
\[
\frac{d}{dx}(x^2y^3) = \frac{d}{dx}(x^2) \cdot y^3 + x^2 \cdot \frac{d}{dx}(y^3)
\]
Differentiating gives:
\[
2x \cdot y^3 + x^2 \cdot 3y^2 \frac{dy}{dx}
\]
This simplifies to:
\[
2x y^3 + 3x^2 y^2 \frac{dy}{dx}
\]

### Step 2: Differentiate the second term \( 4xy \)
Again using the product rule:
\[
\frac{d}{dx}(4xy) = \frac{d}{dx}(4x) \cdot y + 4x \cdot \frac{d}{dx}(y)
\]
Differentiating gives:
\[
4y + 4x \frac{dy}{dx}
\]

### Step 3: Differentiate the third term \( x \)
\[
\frac{d}{dx}(x) = 1
\]

### Step 4: Differentiate the fourth term \( -6y \)
Since \( y \) is a function of \( x \), we apply the chain rule:
\[
\frac{d}{dx}(-6y) = -6 \frac{dy}{dx}
\]

### Step 5: Differentiate the constant on the right-hand side
\[
\frac{d}{dx}(2) = 0
\]

### Step 6: Combine all differentiated terms
Now, combining everything:

\[
(2x y^3 + 3x^2 y^2 \frac{dy}{dx}) + (4y + 4x \frac{dy}{dx}) + 1 - 6 \frac{dy}{dx} = 0
\]

### Step 7: Solve for \( \frac{dy}{dx} \)
Group the terms involving \( \frac{dy}{dx} \):

\[
3x^2 y^2 \frac{dy}{dx} + 4x \frac{dy}{dx} - 6 \frac{dy}{dx} = - (2x y^3 + 4y + 1)
\]

Factor out \( \frac{dy}{dx} \):

\[
\frac{dy}{dx}(3x^2 y^2 + 4x - 6) = -(2x y^3 + 4y + 1)
\]

Finally, solve for \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{-(2x y^3 + 4y + 1)}{3x^2 y^2 + 4x - 6}
\]

This is the derivative \( \frac{dy}{dx} \).

---

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

Here are 5 related questions for further practice:

1. How do you apply implicit differentiation when both variables are involved in trigonometric functions?
2. Can you explain how to use implicit differentiation on the equation \( \sin(xy) + x^2 = y \)?
3. What happens if you differentiate \( x^2 + y^2 = 25 \) implicitly?
4. How would you solve an implicit differentiation problem if it involves higher powers of \( y \)?
5. How do you interpret the result of implicit differentiation geometrically?

**Tip:** When using implicit differentiation, always treat \( y \) as a function of \( x \), and apply the product rule and chain rule where needed!

Learn how to solve the implicit differentiation problem for the equation x^2y^3 + 4xy + x - 6y = 2. This solution covers key calculus concepts such as the product rule and chain rule, suitable for students in Grades 11-12 or 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