Question

Find the derivative of the trigonometric function.
y=2xsinx+x^2cosx

Answer 1

\[y'=\frac{d}{dx}(2xsinx)+ \frac{d}{dx}x^2cosx \]

Answer 2

use product rule

Answer 3

product rule for addition?

Answer 4

use product rule for each term

Answer 5

ah ok

Answer 6

no - apply product rule to each of the two terms
- looks like diyad is doing that now

Answer 7

First do \( \frac{d}{dx}(2xsinx) \) & then \( \frac{d}{dx}(x^2cosx) \)

Answer 8

do you know how to do derivative of 2xsinx ?

Answer 9

d (uv)dx = u* dv/dx + v* du/dx where u and v are function of x

Answer 10

so

Answer 11

2xsix+x^2cosx=2sinx+3xcosx+2xcosx-x^2sinx

Answer 12

2sinx oops

Answer 13

Nope

Answer 14

2xsinx

Answer 15

4xcosx+2sinx-x^2sinx

Answer 16

4xcosx+sinx(2-x^2)

Answer 17

i mistyped a bunch at first lol

Answer 18

\[\frac{d}{dx} (2xsinx)= 2 \frac{d}{dx}(xsinx) =2[x \frac{d}{dx}sinx+sinx \frac{d}{dx}x]\]

Answer 19

my last answer is correct

Answer 20

4xcosx+sinx(2-x^2)

Answer 21

atleast thats what the books says in the back.

Answer 22

\[ \frac{d}{dx}(2xsinx)=2[xcosx+sinx]\]

Answer 23

y=2xsinx+x^2cosx
= 2sinx+2xcosx+2xcosx-x^2sinx
= 4xcosx +2sinx - x^2sinx
= 4xcosx + sinx(2-x^2)

Answer 24

thats correct ChrisV

Answer 25

thanks guys

Answer 26

oh sorry i thought you were taling about the derivative of 2xsinx

Answer 27

did not know i could use product rule on both sides

Answer 28

better to learn than wonder :)

Answer 29

yep - whenever you have a product of two functions the rule applies