find $\frac{dy}{dx}$ by implicit differentiation. $sqrt{x + y}=x^{3}+y^{3}$ $\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the left - hand side

Rewrite $\sqrt{x + y}=(x + y)^{\frac{1}{2}}$. Then differentiate both sides of the equation $(x + y)^{\frac{1}{2}}=x^{3}+y^{3}$ with respect to $x$.

Step2: Differentiate the left - hand side

Using the chain - rule, if $u=x + y$, then $\frac{d}{dx}(x + y)^{\frac{1}{2}}=\frac{1}{2}(x + y)^{-\frac{1}{2}}(1+\frac{dy}{dx})$.

Step3: Differentiate the right - hand side

$\frac{d}{dx}(x^{3}+y^{3}) = 3x^{2}+3y^{2}\frac{dy}{dx}$.

Step4: Set up the equation

We have $\frac{1}{2\sqrt{x + y}}(1+\frac{dy}{dx})=3x^{2}+3y^{2}\frac{dy}{dx}$.

Step5: Expand the left - hand side

$\frac{1}{2\sqrt{x + y}}+\frac{1}{2\sqrt{x + y}}\frac{dy}{dx}=3x^{2}+3y^{2}\frac{dy}{dx}$.

Step6: Isolate $\frac{dy}{dx}$ terms

$\frac{1}{2\sqrt{x + y}}\frac{dy}{dx}-3y^{2}\frac{dy}{dx}=3x^{2}-\frac{1}{2\sqrt{x + y}}$.

Step7: Factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(\frac{1}{2\sqrt{x + y}}-3y^{2})=3x^{2}-\frac{1}{2\sqrt{x + y}}$.

Step8: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{3x^{2}-\frac{1}{2\sqrt{x + y}}}{\frac{1}{2\sqrt{x + y}}-3y^{2}}=\frac{6x^{2}\sqrt{x + y}-1}{1 - 6y^{2}\sqrt{x + y}}$.

Answer:

$\frac{6x^{2}\sqrt{x + y}-1}{1 - 6y^{2}\sqrt{x + y}}$