Find dy/dx y= x sin x How would i do this?

Answer 1

product rule
dy/dx = x cosx + sinx
,

Answer 2

If you integrate both sides,

int y dy/dx dx = int x sin(x) dx,

The integral of the left hand side is 1/2 y^2.

The integral of the right hand side is sin(x)-xcos(x)+C where C is any constant.

Setting them equal you get

1/2 y^2 = sin(x) - xcos(x) + C

Solving for y we get

y = (+/-) sqrt( 2*(sin(x)-xcos(x)+C))

You have to be given initial conditions to determine if you should use the + or the -

Answer 3

You simply use the product rule:
where the product rule is as follows:
the derivative of f(x)g(x) w.r.t. x is:
[df(x)/dx]*g(x) + f(x)*[dg(x)/dx]
hence:
let f(x) = x and g(x) = sin(x)
then df(x)/dx = 1 and dg(x)/dx = cos(x) (as you'll find proven in many calculus textbooks)
Therefore, dy/dx = [df(x)/dx]*g(x) + f(x)*[dg(x)/dx]
= [1]*sin(x) + x*[cos(x)]
= sin(x) + x*cos(x)