anonymous
  • anonymous
Find the derivative of the trigonometric function. y=2xsinx+x^2cosx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Diyadiya
  • Diyadiya
\[y'=\frac{d}{dx}(2xsinx)+ \frac{d}{dx}x^2cosx \]
cwrw238
  • cwrw238
use product rule
anonymous
  • anonymous
product rule for addition?

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More answers

Diyadiya
  • Diyadiya
use product rule for each term
anonymous
  • anonymous
ah ok
cwrw238
  • cwrw238
no - apply product rule to each of the two terms - looks like diyad is doing that now
Diyadiya
  • Diyadiya
First do \( \frac{d}{dx}(2xsinx) \) & then \( \frac{d}{dx}(x^2cosx) \)
Diyadiya
  • Diyadiya
do you know how to do derivative of 2xsinx ?
cwrw238
  • cwrw238
d (uv)dx = u* dv/dx + v* du/dx where u and v are function of x
anonymous
  • anonymous
so
anonymous
  • anonymous
2xsix+x^2cosx=2sinx+3xcosx+2xcosx-x^2sinx
anonymous
  • anonymous
2sinx oops
Diyadiya
  • Diyadiya
Nope
anonymous
  • anonymous
2xsinx
anonymous
  • anonymous
4xcosx+2sinx-x^2sinx
anonymous
  • anonymous
4xcosx+sinx(2-x^2)
anonymous
  • anonymous
i mistyped a bunch at first lol
Diyadiya
  • Diyadiya
\[\frac{d}{dx} (2xsinx)= 2 \frac{d}{dx}(xsinx) =2[x \frac{d}{dx}sinx+sinx \frac{d}{dx}x]\]
anonymous
  • anonymous
my last answer is correct
anonymous
  • anonymous
4xcosx+sinx(2-x^2)
anonymous
  • anonymous
atleast thats what the books says in the back.
Diyadiya
  • Diyadiya
\[ \frac{d}{dx}(2xsinx)=2[xcosx+sinx]\]
anonymous
  • anonymous
y=2xsinx+x^2cosx = 2sinx+2xcosx+2xcosx-x^2sinx = 4xcosx +2sinx - x^2sinx = 4xcosx + sinx(2-x^2)
cwrw238
  • cwrw238
thats correct ChrisV
anonymous
  • anonymous
thanks guys
Diyadiya
  • Diyadiya
oh sorry i thought you were taling about the derivative of 2xsinx
anonymous
  • anonymous
did not know i could use product rule on both sides
anonymous
  • anonymous
better to learn than wonder :)
cwrw238
  • cwrw238
yep - whenever you have a product of two functions the rule applies

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