To find the derivative of the equation \(x^4 + \sin y = x^3 y^2\) with respect to \(x\), we will use implicit differentiation.
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Differentiate both sides of the equation with respect to \(x\):
- The derivative of \(x^4\) is \(4x^3\).
- The derivative of \(\sin y\) with respect to \(x\) is \(\cos y \frac{dy}{dx}\) (using the chain rule).
- The derivative of \(x^3 y^2\) is \(3x^2 y^2 + x^3 \cdot 2y \frac{dy}{dx}\) (using the product rule).
Putting it all together, we differentiate both sides:
\[ \frac{d}{dx}(x^4) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(x^3 y^2) \]
This results in:
\[ 4x^3 + \cos y \frac{dy}{dx} = 3x^2 y^2 + x^3 \cdot 2y \frac{dy}{dx} \]
- Rearrange the equation to solve for \(\frac{dy}{dx}\):
Combine all the terms involving \(\frac{dy}{dx}\):
\[ \cos y \frac{dy}{dx} - 2x^3 y \frac{dy}{dx} = 3x^2 y^2 - 4x^3 \]
Factor out \(\frac{dy}{dx}\):
\[ \left(\cos y - 2x^3 y\right) \frac{dy}{dx} = 3x^2 y^2 - 4x^3 \]
- Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{3x^2 y^2 - 4x^3}{\cos y - 2x^3 y} \]
Thus, the derivative of the equation \(x^4 + \sin y = x^3 y^2\) with respect to \(x\) is:
\[ \frac{dy}{dx} = \frac{3x^2 y^2 - 4x^3}{\cos y - 2x^3 y} \]