answersLogoWhite

0

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

User Avatar

Wiki User

15y ago

Still curious? Ask our experts.

Chat with our AI personalities

ProfessorProfessor
I will give you the most educated answer.
Chat with Professor
DevinDevin
I've poured enough drinks to know that people don't always want advice—they just want to talk.
Chat with Devin
ReneRene
Change my mind. I dare you.
Chat with Rene
More answers

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)

User Avatar

Wiki User

12y ago
User Avatar

Add your answer:

Earn +20 pts
Q: What is the derivative of x 2cosx?
Write your answer...
Submit
Still have questions?
magnify glass
imp