QuestionGiven $\sqrt{x+y}=y^{3}$, find $\frac{d y}{d x}$ in terms of $x$ and $y$.

Studdy Solution
Solve for $\frac{dy}{dx}$.
First, distribute on the left side: $$\frac{1}{2\sqrt{x+y}} + \frac{1}{2\sqrt{x+y}} \cdot \frac{dy}{dx} = 3y^2 \cdot \frac{dy}{dx}$$
Rearrange terms to isolate $\frac{dy}{dx}$: $$\frac{1}{2\sqrt{x+y}} = 3y^2 \cdot \frac{dy}{dx} - \frac{1}{2\sqrt{x+y}} \cdot \frac{dy}{dx}$$
Factor out $\frac{dy}{dx}$ from the right side: $$\frac{1}{2\sqrt{x+y}} = \left(3y^2 - \frac{1}{2\sqrt{x+y}}\right) \cdot \frac{dy}{dx}$$
Finally, solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\frac{1}{2\sqrt{x+y}}}{3y^2 - \frac{1}{2\sqrt{x+y}}}$$
The derivative $\frac{dy}{dx}$ in terms of $x$ and $y$ is:
$$\frac{dy}{dx} = \frac{1}{2\sqrt{x+y} \left(3y^2 - \frac{1}{2\sqrt{x+y}}\right)}$$