Question

use implicit differentiation to find $\frac{dy}{dx}$.
$(4xy + 3)^2=16y$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate \((4xy + 3)^2\) using chain - rule and product - rule. The derivative of the left - hand side: Let \(u = 4xy+3\), then \((4xy + 3)^2=u^2\). By the chain - rule, \(\frac{d}{dx}(u^2)=2u\frac{du}{dx}\). And \(\frac{du}{dx}=4y + 4x\frac{dy}{dx}\). So \(\frac{d}{dx}(4xy + 3)^2=2(4xy + 3)(4y + 4x\frac{dy}{dx})\). The derivative of the right - hand side is \(16\frac{dy}{dx}\). So we have \(2(4xy + 3)(4y + 4x\frac{dy}{dx})=16\frac{dy}{dx}\).

Step2: Expand the left - hand side

Expand \(2(4xy + 3)(4y + 4x\frac{dy}{dx})\):
\[

$$\begin{align*} 2(4xy + 3)(4y + 4x\frac{dy}{dx})&=2(16xy^{2}+16x^{2}y\frac{dy}{dx}+12y + 12x\frac{dy}{dx})\\ &=32xy^{2}+32x^{2}y\frac{dy}{dx}+24y + 24x\frac{dy}{dx} \end{align*}$$

\]
So \(32xy^{2}+32x^{2}y\frac{dy}{dx}+24y + 24x\frac{dy}{dx}=16\frac{dy}{dx}\).

Step3: Isolate \(\frac{dy}{dx}\) terms

Move all terms with \(\frac{dy}{dx}\) to one side:
\[32x^{2}y\frac{dy}{dx}+24x\frac{dy}{dx}-16\frac{dy}{dx}=-32xy^{2}-24y\]
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(32x^{2}y + 24x-16)=-32xy^{2}-24y\).

Step4: Solve for \(\frac{dy}{dx}\)

\(\frac{dy}{dx}=\frac{-32xy^{2}-24y}{32x^{2}y + 24x - 16}=\frac{-8xy^{2}-6y}{8x^{2}y + 6x - 4}\)

Answer:

\(\frac{-8xy^{2}-6y}{8x^{2}y + 6x - 4}\)