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

RossRoss
Every question is just a happy little opportunity.
Chat with Ross
ViviVivi
Your ride-or-die bestie who's seen you through every high and low.
Chat with Vivi
EzraEzra
Faith is not about having all the answers, but learning to ask the right questions.
Chat with Ezra
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

11y 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