### Step-by-Step Derivative Solution**Part 2 of 5**On the left-hand side, for the first term \(9x^3\), we have:\[\frac{d}{dx}[9x^3] = 27x^2\]\[ \text{Checked: } 27x^2 \]Now, note that the second term \(x^2y\) is a product, and so we must use the Product Rule. We have:\[\frac{d}{dx}[x^2y] = (x^2)(y') + (y)(2x)\]\[ \text{Checked: } 2x \]**Part 3 of 5**For the last term \(xy^3\), note that it is also a product, so we must again use the Product Rule. We, therefore, have:\[\frac{d}{dx}[xy^3] = (x)(3y^2)(y') + (y^3)(1)\]\[ \text{Checked: } 1 \]**Part 4 of 5**Substituting the above back into \[\frac{d}{dx}[9x^3 + x^2y - xy^3] = \frac{d}{dx}[1],\]we now have the equation:\[27x^2 + (x^2y' + 2xy) - (3xy^2y' + y^3) = 0.\]We rearrange so that all terms containing \(y'\) are on the same side, and get:\[27x^2 + 2xy - y^3 = 3xy^2y' - x^2y'.\]\[ \text{Checked: } 3xy^2 - x^2 \]**Part 5 of 5**Finally, we factor out the \(y'\) and divide to conclude that the derivative is:\[ y' = \frac{27x^2 + 2xy - y^3}{3xy^2 - x^2}. \]

Name: ### Step-by-Step Derivative Solution**Part 2 of 5**On the left-hand s
Uploaded: 2024-03-20T06:47:22.000Z
Channel: Bartleby
Description: ### Step-by-Step Derivative Solution**Part 2 of 5**On the left-hand side, for the first term \(9x^3\), we have:\[\frac{d}{dx}[9x^3] = 27x^2\]\[ \text{Checked: } 27x^2 \]Now, note that the second term \(x^2y\) is a product, and so we must use the Pro