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

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

Mathematics
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\[y'=\frac{d}{dx}(2xsinx)+ \frac{d}{dx}x^2cosx \]
use product rule
product rule for addition?

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Other answers:

use product rule for each term
ah ok
no - apply product rule to each of the two terms - looks like diyad is doing that now
First do \( \frac{d}{dx}(2xsinx) \) & then \( \frac{d}{dx}(x^2cosx) \)
do you know how to do derivative of 2xsinx ?
d (uv)dx = u* dv/dx + v* du/dx where u and v are function of x
so
2xsix+x^2cosx=2sinx+3xcosx+2xcosx-x^2sinx
2sinx oops
Nope
2xsinx
4xcosx+2sinx-x^2sinx
4xcosx+sinx(2-x^2)
i mistyped a bunch at first lol
\[\frac{d}{dx} (2xsinx)= 2 \frac{d}{dx}(xsinx) =2[x \frac{d}{dx}sinx+sinx \frac{d}{dx}x]\]
my last answer is correct
4xcosx+sinx(2-x^2)
atleast thats what the books says in the back.
\[ \frac{d}{dx}(2xsinx)=2[xcosx+sinx]\]
y=2xsinx+x^2cosx = 2sinx+2xcosx+2xcosx-x^2sinx = 4xcosx +2sinx - x^2sinx = 4xcosx + sinx(2-x^2)
thats correct ChrisV
thanks guys
oh sorry i thought you were taling about the derivative of 2xsinx
did not know i could use product rule on both sides
better to learn than wonder :)
yep - whenever you have a product of two functions the rule applies

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