What is the derivative of x 2cosx?

Answer 1

For f(x) + g(x), the derivative d/dx[f(x) + g(x)] = f'(x) + g'(x). Essentially, this just means that since it's addition, you can take the derivative of each part.

d/dx(x - 2cosx) =

* d/dx is a way to indicate you're taking the derivative

d/dx(x) + d/dx(-2cosx)

* take the derivative of each part, distributing

(1) + (-2*-sinx)

* f(x) = x, f'(x) = 1 & g(x) = -2cosx, g'(x) = +2sinx

= 1 + 2sinx

Answer 2

I'll solve two derivatives for you, because I'm not sure if you meant x squared or 2x.

y = x2cos(x)y` = ?

We need to use the product rule:

If a = x2 and b = cos(x) then:
y` = a`b + ab`

a` = 2x

b` = -sin(x)

y` = (2x)cos(x)-x2sin(x)

y = 2xcos(x)y` = ?

Again we need to use the product rule:

a = 2x, b = cos(x)

a`= 2, b`= -sin(x)

y` = a`b + ab`

y` = 2cos(x)-(2x)sin(x)

Once again, I solved two different problems here, because I was unsure about what problem you were asking about:

[x2cos(x)]` =(2x)cos(x)-x2sin(x)

[2xcos(x)]` =2cos(x)-(2x)sin(x)