Question

Find the derivative of x⁴+sin y=x³y²

Answer 1

To find the derivative of the equation \(x^4 + \sin y = x^3 y^2\) with respect to \(x\), we will use implicit differentiation.

1. 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:

\[
4x^3 + \cos y \frac{dy}{dx} = 3x^2 y^2 + x^3 \cdot 2y \frac{dy}{dx}
\]

2. 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 \(\frac{dy}{dx}\):

\[
\left(\cos y - 2x^3 y\right) \frac{dy}{dx} = 3x^2 y^2 - 4x^3
\]

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:

\[
\frac{dy}{dx} = \frac{3x^2 y^2 - 4x^3}{\cos y - 2x^3 y}
\]