Answer:
The derivative of yS(siny + ycosy)dy with respect to y is yS(cosy - ysiny) + S(siny + ycosy).
Step-by-step explanation:
To differentiate the expression yS(siny + y cosy)dy, we need to apply the product rule of differentiation. The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
(d/dx)(u(x) v(x)) = u'(x) v(x) + u(x) v'(x)
Using this rule, we can differentiate the expression as follows:
Let u(y) = yS and v(y) = siny + ycosy. Then,
u'(y) = S (the derivative of yS with respect to y is S)
v'(y) = cosy + y(-siny) = cosy - ysiny (using the product rule again)
Using the product rule, we have:
(d/dy)(u(y) v(y)) = u'(y) v(y) + u(y) v'(y)
= yS (cosy - ysiny) + S(siny + ycosy)
Therefore, the derivative of yS(siny + ycosy)dy with respect to y is yS(cosy - ysiny) + S(siny + ycosy).