Question

given 4cos(2x)sin(y)=7, find dy/dx by implicit differentiation

Answer 1

\[4\cos(2x)\sin y=7\]
Differentiating both sides w.r.t. x, you have
\[\begin{align*}\frac{d}{dx}[4\cos(2x)\sin y]&=\frac{d}{dx}[7]\\
4\frac{d}{dx}[\cos(2x)\sin y]&=0\\
4\left(\frac{d}{dx}[\cos(2x)]\sin y+\cos(2x)\frac{d}{dx}[\sin y]\right)&=0\\
4\left(-\sin(2x)\frac{d}{dx}[2x]\sin y+\cos(2x)\cos y\frac{d}{dx}[y]\right)&=0\\
4\left(-2\sin(2x)\sin y+\cos(2x)\cos y\frac{dy}{dx}\right)&=0\end{align*}\]
I'm sure you can take it from here. I basically just applied the chain and product rules to get terms containing dy/dx.

Answer 2

Now what do I do?

Answer 3

Solve for \(\displaystyle\frac{dy}{dx}\). Distribute the 4, move all terms with \(\displaystyle\frac{dy}{dx}\) to one side, then divide by the terms in front of it. Basic algebra.