Final answer:
To find the differential of v = 3y cos(xy), we use the product rule combined with the chain rule. The result is dv = (3 cos(xy) - 3yx sin(xy))dy + (-3y sin(xy))x dx.
Step-by-step explanation:
To find the differential of the function v = 3y cos(xy), we apply the product rule and chain rule of differentiation. The product rule is expressed as d(uv) = udv + vdu, where u and v are functions of y. In this case, we treat 3y as u and cos(xy) as v.
The differential udv is calculated as 3y times the differential of cos(xy), which requires the chain rule since xy is a function of y. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function. So, the differential of cos(xy) would be -sin(xy) times the derivative of xy, which is x.
Therefore, the differential udv is -3yx sin(xy)dx. Similarly, the differential vdu is the derivative of 3y, which is 3, times cos(xy). Hence, vdu = 3 cos(xy)dy.
Combining these two differentials, we get:
dv = udv + vdu
So:
dv = (3 cos(xy) - 3yx sin(xy))dy + (-3y sin(xy))x dx
When separating the variables, we place all the terms including v and dv on one side, and all terms including y and dy on the other side, which is shown as ady = udv in your reference material.