1. Start with the equation: \( e^{2x - 9y} + x^3 y^2 = \sin(y) + 11x \).
2. Differentiate both sides with respect to \( x \), using the chain rule and product rule where necessary.
3. Differentiate \( e^{2x - 9y} \):
- Use the chain rule: \( \frac{d}{dx} e^{2x - 9y} = e^{2x - 9y} \cdot \frac{d}{dx}(2x - 9y) \).
- \( \frac{d}{dx}(2x - 9y) = 2 - 9y' \).
- So, \( \frac{d}{dx} e^{2x - 9y} = e^{2x - 9y} (2 - 9y') \).
4. Differentiate \( x^3 y^2 \):
- Use the product rule: \( \frac{d}{dx}(x^3 y^2) = 3x^2 y^2 + x^3 \cdot 2y y' \).
- So, \( \frac{d}{dx}(x^3 y^2) = 3x^2 y^2 + 2x^3 y y' \).
5. Differentiate \( \sin(y) \):
- Use the chain rule: \( \frac{d}{dx} \sin(y) = \cos(y) \cdot y' \).
6. Differentiate \( 11x \):
- \( \frac{d}{dx} 11x = 11 \).
7. Combine all terms:
- \( e^{2x - 9y} (2 - 9y') + 3x^2 y^2 + 2x^3 y y' = \cos(y) y' + 11 \).
8. Collect terms involving \( y' \):
- \( e^{2x - 9y} \cdot 2 - 9 e^{2x - 9y} y' + 3x^2 y^2 + 2x^3 y y' = \cos(y) y' + 11 \).
- Move all terms with \( y' \) to one side and others to the opposite side:
- \( -9 e^{2x - 9y} y' + 2x^3 y y' - \cos(y) y' = 11 - 2 e^{2x - 9y} - 3x^2 y^2 \).
9. Factor out \( y' \):
- \( y' (-9 e^{2x - 9y} + 2x^3 y - \cos(y)) = 11 - 2 e^{2x - 9y} - 3x^2 y^2 \).
10. Solve for \( y' \):
- \( y' = \frac{11 - 2 e^{2x - 9y} - 3x^2 y^2}{-9 e^{2x - 9y} + 2x^3 y - \cos(y)} \).
Final Answer:
\( y' = \frac{11 - 2 e^{2x - 9y} - 3x^2 y^2}{-9 e^{2x - 9y} + 2x^3 y - \cos(y)} \)
Parent Tip: Review the logic above to help your child master the concept of derivative cheat sheet.
