Find dy/dx by implicit differentiation.
x cos y + y cos x = 1
just use the product rule and go for it:
cosy - x siny y' + y' cosx - y sinx = 0
y'(cosx - x siny) = y sinx - cosy
y' = (y sinx - cosy) / (cosx - x siny)
To find dy/dx by implicit differentiation, we will differentiate both sides of the equation with respect to x while treating y as an implicit function of x.
Let's go step by step:
Step 1: Differentiate both sides of the equation with respect to x using the product rule and chain rule as necessary. When differentiating y with respect to x, we treat y as a function of x.
On the left side:
Differentiate x cos y with respect to x using the product rule:
d/dx (x cos y) = (1)(cos y) + (x)(-sin y)(dy/dx) = cos y - x sin y (dy/dx)
On the right side:
Differentiate 1 with respect to x, and since 1 is a constant, its derivative is 0.
Therefore, our equation becomes:
(cos y - x sin y)(dy/dx) + y(-sin x) = 0
Step 2: Solve the equation for dy/dx.
Rearrange the equation by isolating dy/dx:
(cos y - x sin y)(dy/dx) = -y(-sin x)
(dy/dx) = -y(-sin x) / (cos y - x sin y)
And that is the expression for dy/dx in terms of x and y.
Note: If you need to simplify this expression further, you can manipulate it algebraically by factoring out common terms or rearranging the terms.