Question

use implicit differentiation to determine dy/dx of xy=sinx + siny

Answer 1

xy = sinx + siny y + x dy/dx = cosx + cosy dy/dx dy/dx (x - cosy) = cosx - y dy/dx = (cosx - y) / (x - cosy)

Answer 2

understand what I did?

Answer 3

no, very new at implicit, and would like to see someone work through the steps

Answer 4

\[xy = \sin(x)\]\[\frac{d}{dx}(xy = \sin(x))\]\[\sf \text{product rule} : f'g+g'f \]When using the product rule, \(f =x\) and \(g=y\), when you take the derivative of g, you are taking the derivative of y, w.r.t x, or written as \(\dfrac{dy}{dx}\) \[1\cdot y+\frac{dy}{dx}\cdot x = \cos(x)\]

Answer 5

crud, I wrote it wrong, I revised the question...sorry