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What is the derivative of #y=xcosx+sinx#?

Answer 1

#2cos x-x*sinx#

derrivtive of #x##cosx# can be fund bu uv rule and is #cosx-x*sinx# and #sinx# derivative is cos x adding both answers lead to #2cosx-x*sinx#

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Answer 2

Paisley Johnson

To find the derivative of ( y = x \cos(x) + \sin(x) ), you can use the product rule and the derivatives of cosine and sine functions.

[ \frac{d}{dx} (x \cos(x) + \sin(x)) = \frac{d}{dx}(x) \cdot \cos(x) + x \cdot \frac{d}{dx}(\cos(x)) + \frac{d}{dx}(\sin(x)) ]

Using the product rule (\frac{d}{dx}(uv) = u'v + uv'), where (u = x) and (v = \cos(x)), and the derivatives of (\cos(x)) and (\sin(x)):

[ \frac{d}{dx}(x \cos(x) + \sin(x)) = 1 \cdot \cos(x) + x \cdot (-\sin(x)) + \cos(x) ]

[ = \cos(x) - x \sin(x) + \cos(x) ]

So, the derivative of ( y = x \cos(x) + \sin(x) ) is ( \cos(x) - x \sin(x) + \cos(x) ), which simplifies to ( 2 \cos(x) - x \sin(x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.