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

Answer 1

Quotient rule states that for a function #y=f(x)/g(x)#, then #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2#

Resolution:

#(dy)/(dx)=(e^x(x-1)-e^x(1))/(x-1)^2#
#(dy)/(dx)=(e^x(x-2))/(x-1)^2#
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Answer 2

To differentiate ( \frac{e^x}{x-1} ) using the quotient rule:

Let ( u = e^x ) and ( v = x - 1 ).

Then, ( u' = e^x ) and ( v' = 1 ).

According to the quotient rule:

[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ]

Substitute the values:

[ \frac{d}{dx}\left(\frac{e^x}{x-1}\right) = \frac{(e^x)(x - 1) - (e^x)(1)}{(x - 1)^2} ]

[ = \frac{e^x(x - 1) - e^x}{(x - 1)^2} ]

[ = \frac{e^x(x - 1 - 1)}{(x - 1)^2} ]

[ = \frac{e^x(x - 2)}{(x - 1)^2} ]

So, the derivative of ( \frac{e^x}{x-1} ) using the quotient rule is ( \frac{e^x(x - 2)}{(x - 1)^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.

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.

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.

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.

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