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How do you find the derivative of #y= (x+1) / (x-1)#?

Answer 1

Christopher Agresta

#= -( 2 )/((x-1)^2)#

Quotient Rule

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

here #u = x+1, v = x-1, u' = v' = 1#

so we have

# ((1) (x-1) - (x+1) (1))/((x-1)^2)#

#= -( 2 )/((x-1)^2)#

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Answer 2

Charles Arocha

To find the derivative of ( y = \frac{x+1}{x-1} ), you would use the quotient rule:

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

In this case, let ( u = x + 1 ) and ( v = x - 1 ). Then, differentiate ( u ) and ( v ) with respect to ( x ) to find ( u' ) and ( v' ):

[ u' = 1 ] [ v' = 1 ]

Now, substitute these values into the quotient rule formula:

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

Simplify the numerator and denominator:

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

So, the derivative of ( y = \frac{x+1}{x-1} ) is ( \frac{-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.