en the equation \( \tan x=5 \sin y-4 x y \), find \( \frac{d y}{d x} \) by implicit differentiation.

To find \( \frac{dy}{dx} \) by implicit differentiation for the equation \( \tan x = 5 \sin y - 4xy \), we will differentiate both sides with respect to \( x \). Starting with the left-hand side: \[ \frac{d}{dx}(\tan x) = \sec^2 x. \] Now we differentiate the right-hand side \( 5 \sin y - 4xy \) using the product and chain rules: \[ \frac{d}{dx}(5 \sin y) = 5 \cos y \frac{dy}{dx}, \] and \[ \frac{d}{dx}(-4xy) = -4\left(x \frac{dy}{dx} + y\right). \] Putting it all together, we have: \[ \sec^2 x = 5 \cos y \frac{dy}{dx} - 4\left(x \frac{dy}{dx} + y\right). \] Now, distributing and rearranging gives: \[ \sec^2 x = 5 \cos y \frac{dy}{dx} - 4x \frac{dy}{dx} - 4y. \] Combine the terms involving \( \frac{dy}{dx} \): \[ \sec^2 x + 4y = \left(5 \cos y - 4x\right) \frac{dy}{dx}. \] Now, we isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\sec^2 x + 4y}{5 \cos y - 4x}. \] Thus, \[ \frac{dy}{dx} = \frac{\sec^2 x + 4y}{5 \cos y - 4x}. \]