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



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

cwrw238
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use product rule

ChrisV
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product rule for addition?

Diyadiya
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use product rule for each term

ChrisV
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ah ok

cwrw238
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no  apply product rule to each of the two terms
 looks like diyad is doing that now

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

Diyadiya
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do you know how to do derivative of 2xsinx ?

cwrw238
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d (uv)dx = u* dv/dx + v* du/dx where u and v are function of x

ChrisV
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so

ChrisV
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2xsix+x^2cosx=2sinx+3xcosx+2xcosxx^2sinx

ChrisV
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2sinx oops

Diyadiya
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Nope

ChrisV
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2xsinx

ChrisV
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4xcosx+2sinxx^2sinx

ChrisV
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4xcosx+sinx(2x^2)

ChrisV
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i mistyped a bunch at first lol

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

ChrisV
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my last answer is correct

ChrisV
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4xcosx+sinx(2x^2)

ChrisV
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atleast thats what the books says in the back.

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

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

cwrw238
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thats correct ChrisV

ChrisV
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thanks guys

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

ChrisV
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did not know i could use product rule on both sides

ChrisV
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better to learn than wonder :)

cwrw238
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yep  whenever you have a product of two functions the rule applies