Question

use implicit differentiation to find $\frac{dy}{dx}$ for the equation $\sqrt{x + y}=8 + x^{2}y^{2}$. $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate \(\sqrt{x + y}\) and \(8+x^{2}y^{2}\) with respect to \(x\).
For the left - hand side, using the chain rule, if \(u=x + y\), then \(\frac{d}{dx}(\sqrt{u})=\frac{1}{2\sqrt{u}}\cdot(1 + \frac{dy}{dx})\). So \(\frac{d}{dx}(\sqrt{x + y})=\frac{1 + \frac{dy}{dx}}{2\sqrt{x + y}}\).
For the right - hand side, using the sum rule and the product rule. The derivative of a constant \(8\) is \(0\). For \(x^{2}y^{2}\), by the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = x^{2}\), \(u^\prime=2x\) and \(v = y^{2}\), \(v^\prime = 2y\frac{dy}{dx}\). So \(\frac{d}{dx}(x^{2}y^{2})=2xy^{2}+2x^{2}y\frac{dy}{dx}\).
We get \(\frac{1+\frac{dy}{dx}}{2\sqrt{x + y}}=2xy^{2}+2x^{2}y\frac{dy}{dx}\).

Step2: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(2\sqrt{x + y}\) to get \(1+\frac{dy}{dx}=4xy^{2}\sqrt{x + y}+4x^{2}y\sqrt{x + y}\frac{dy}{dx}\).
Rearrange terms: \(\frac{dy}{dx}-4x^{2}y\sqrt{x + y}\frac{dy}{dx}=4xy^{2}\sqrt{x + y}-1\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(1 - 4x^{2}y\sqrt{x + y})=4xy^{2}\sqrt{x + y}-1\).
Then \(\frac{dy}{dx}=\frac{4xy^{2}\sqrt{x + y}-1}{1 - 4x^{2}y\sqrt{x + y}}\).

Answer:

\(\frac{4xy^{2}\sqrt{x + y}-1}{1 - 4x^{2}y\sqrt{x + y}}\)