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How do you differentiate #f(x)=xe^x-x# using the product rule?

Answer 1

William Abdalla

#f'(x)=e^x+xe^x-1#

Let's first examine just the #xe^x# portion of this function.

Through the product rule, #d/dx[xe^x]=color(blue)(d/dx[x])*e^x+color(purple)(d/dx[e^x]*)x#

Therefore, #d/dx[xe^x]=color(blue)(1)*e^x+color(purple)(e^x)*x=e^x+xe^x#

We can use this in the original equation to determine that #f'(x)=e^x+xe^x-d/dx[x]=color(green)(e^x+xe^x-1#

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Answer 2

Aurora Alvarado

To differentiate ( f(x) = xe^x - x ) using the product rule, we differentiate each term separately and then apply the product rule, which states that if ( f(x) = u(x)v(x) ), then ( f'(x) = u'(x)v(x) + u(x)v'(x) ).

Here's how we do it:

Let ( u(x) = x ) and ( v(x) = e^x ).
Find ( u'(x) ) and ( v'(x) ). ( u'(x) = 1 ) (derivative of ( x )) ( v'(x) = e^x ) (derivative of ( e^x ))
Apply the product rule: ( f'(x) = u'(x)v(x) + u(x)v'(x) ) ( = (1)(e^x) + (x)(e^x) ) ( = e^x + xe^x )

So, ( f'(x) = e^x + xe^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.