find $\frac{dy}{dx}$ by implicit differentiation. $sqrt{x + y}=x^{7}+y^{7}$ $\frac{dy}{dx}=$

Explanation:

Step1: Rewrite the left - hand side

Rewrite $\sqrt{x + y}=(x + y)^{\frac{1}{2}}$. Then differentiate both sides of the equation $(x + y)^{\frac{1}{2}}=x^{7}+y^{7}$ with respect to $x$. Using the chain - rule, the derivative of the left - hand side is $\frac{1}{2}(x + y)^{-\frac{1}{2}}(1+\frac{dy}{dx})$. The derivative of the right - hand side is $7x^{6}+7y^{6}\frac{dy}{dx}$. So we have $\frac{1}{2\sqrt{x + y}}(1 + \frac{dy}{dx})=7x^{6}+7y^{6}\frac{dy}{dx}$.

Step2: Expand the left - hand side

Multiply both sides of the equation by $2\sqrt{x + y}$ to get $1+\frac{dy}{dx}=2\sqrt{x + y}(7x^{6}+7y^{6}\frac{dy}{dx})$. Expand the right - hand side: $1+\frac{dy}{dx}=14x^{6}\sqrt{x + y}+14y^{6}\sqrt{x + y}\frac{dy}{dx}$.

Step3: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $\frac{dy}{dx}-14y^{6}\sqrt{x + y}\frac{dy}{dx}=14x^{6}\sqrt{x + y}-1$. Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(1 - 14y^{6}\sqrt{x + y})=14x^{6}\sqrt{x + y}-1$.

Step4: Solve for $\frac{dy}{dx}$

Divide both sides by $(1 - 14y^{6}\sqrt{x + y})$ to get $\frac{dy}{dx}=\frac{14x^{6}\sqrt{x + y}-1}{1 - 14y^{6}\sqrt{x + y}}$.

Answer:

$\frac{14x^{6}\sqrt{x + y}-1}{1 - 14y^{6}\sqrt{x + y}}$