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

\[
x^2y + xy^2 = 6
\]

we will use **implicit differentiation** since the equation involves both \(x\) and \(y\) terms. Here's the process step by step:

### 1. Differentiate both sides with respect to \(x\):
Since \(y\) is a function of \(x\), we apply the product rule and chain rule.

#### Left-hand side:
For the term \(x^2y\):
\[
\frac{d}{dx}(x^2y) = x^2 \frac{dy}{dx} + 2xy
\]
(Using the product rule: \(f'(x)g(x) + f(x)g'(x)\))

For the term \(xy^2\):
\[
\frac{d}{dx}(xy^2) = y^2 + 2xy \frac{dy}{dx}
\]
(Again applying the product rule)

#### Right-hand side:
The derivative of the constant 6 is 0:
\[
\frac{d}{dx}(6) = 0
\]

### 2. Combine all the differentiated terms:
\[
x^2 \frac{dy}{dx} + 2xy + y^2 + 2xy \frac{dy}{dx} = 0
\]

### 3. Group the terms involving \(\frac{dy}{dx}\) together:
\[
x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} = -2xy - y^2
\]

### 4. Factor out \(\frac{dy}{dx}\):
\[
(x^2 + 2xy) \frac{dy}{dx} = - (2xy + y^2)
\]

### 5. Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{-(2xy + y^2)}{x^2 + 2xy}
\]

Thus, the derivative \(\frac{dy}{dx}\) for the curve \(x^2y + xy^2 = 6\) is:
\[
\frac{dy}{dx} = \frac{-(2xy + y^2)}{x^2 + 2xy}
\]

Would you like more details on any part of the solution or further examples?

### Related Questions:
1. How do you apply implicit differentiation to more complex equations?
2. What is the product rule in differentiation, and when is it used?
3. How do you find the tangent line of a curve at a given point?
4. What is the significance of \(\frac{dy}{dx}\) in geometric interpretation?
5. How can you solve higher-order derivatives for implicit functions?

**Tip**: Always simplify as much as possible after applying implicit differentiation to make further steps easier.

Find dy/dx for the curve x^2y + xy^2 = 6.

Learn how to apply implicit differentiation to find dy/dx for the curve x^2y + xy^2 = 6. This example demonstrates the use of the product rule and chain rule, essential for calculus students in Grade 11-12 or first-year college calculus.

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