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How do you differentiate #f(x) = (e^x-1)/(x-e^x)# using the quotient rule?

Answer 1

Christopher Agresta

The answer is #f'(x)=(xe^x-2e^x+1)/(x-e^x)^2#

We use

#(u/v)'=(u'v-uv')/v^2#

Here,

#f(x)=(e^x-1)/(x-e^x)#

#u=e^x-1#, #=>#, #u'=e^x#

#v=x-e^x#, #=>#, #v'=1-e^x#

#f'(x)=(e^x(x-e^x)-(e^x-1)(1-e^x))/(x-e^x)^2#

#=(xe^x-(e^x)^2+(e^x)^2-e^x+1-e^x)/((x-e^x)^2)#

#=(xe^x-2e^x+1)/(x-e^x)^2#

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Answer 2

Cameron Alexander

To differentiate ( f(x) = \frac{e^x - 1}{x - e^x} ) using the quotient rule, follow these steps:

Let ( u(x) = e^x - 1 ) and ( v(x) = x - e^x ).
Find ( u'(x) ) and ( v'(x) ).
- ( u'(x) = e^x ) (derivative of ( e^x - 1 ))
- ( v'(x) = 1 - e^x ) (derivative of ( x - e^x ))
Apply the quotient rule:
- ( f'(x) = \frac{u'(x)v(x) - v'(x)u(x)}{(v(x))^2} )
- ( f'(x) = \frac{(e^x)(x - e^x) - (1 - e^x)(e^x - 1)}{(x - e^x)^2} )
Simplify the expression:
- ( f'(x) = \frac{xe^x - (e^x)^2 - e^x + 1 - e^x + e^{2x}}{(x - e^x)^2} )
- ( f'(x) = \frac{xe^x - e^{2x} - 2e^x + 1}{(x - e^x)^2} )

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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.