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How do you use the product rule to find the derivative of #y=x^2*sin(x)# ?

Answer 1

Scarlett Allison

#y'=x(2sinx+xcosx)#

The Product Rule tells us that if we have two functions #f(x), g(x),# multiplied by one another, then

#(fg)'=fg'+gf'#

Here, for #y=x^2sinx,# we see #f(x)=x^2, g(x)=sinx#

Thus,

#y'=sinxd/dxx^2+x^2d/dxsinx#

#y'=2xsinx+x^2cosx#

Factor out #x# to simplify:

#y'=x(2sinx+xcosx)#

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Answer 2

Isaiah Allen

To find the derivative of ( y = x^2 \cdot \sin(x) ) using the product rule, you would differentiate each term separately and then apply the product rule formula, which states:

[ (f \cdot g)' = f' \cdot g + f \cdot g' ]

Where ( f ) and ( g ) are functions of ( x ). Here, ( f(x) = x^2 ) and ( g(x) = \sin(x) ).

[ f'(x) = 2x ] [ g'(x) = \cos(x) ]

Now apply the product rule formula:

[ y' = (x^2)' \cdot \sin(x) + x^2 \cdot (\sin(x))' ]

[ y' = (2x) \cdot \sin(x) + x^2 \cdot \cos(x) ]

[ y' = 2x \sin(x) + x^2 \cos(x) ]

So, the derivative of ( y = x^2 \cdot \sin(x) ) is ( y' = 2x \sin(x) + x^2 \cos(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.