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How do you find the derivative of #y = (x+2)/(x-2) #?

Answer 1

Charles Arocha

Using the quotient rule, which states that for #y=f(x)/g(x)#, we'll have #y'=(f'(x)g(x)-f(x)g'(x))/(g(x)^2)#, we have

#(dy)/(dx)=(1*(x-2)-(x+2)*1)/(x-2)^2=(cancel(x)-2-cancel(x)-2)/(x-2)^2=-4/(x-2)^2#

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Answer 2

Charles Anctil

To find the derivative of ( y = \frac{x+2}{x-2} ), you can use the quotient rule, which states that if ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ), where ( u ) and ( v ) are functions of ( x ), and ( u' ) and ( v' ) represent their respective derivatives with respect to ( x ).

Let ( u = x + 2 ) and ( v = x - 2 ). Then ( u' = 1 ) and ( v' = 1 ).

Now apply the quotient rule: [ y' = \frac{(1)(x - 2) - (x + 2)(1)}{(x - 2)^2} ]

Simplify: [ y' = \frac{x - 2 - x - 2}{(x - 2)^2} ] [ y' = \frac{-4}{(x - 2)^2} ]

So, the derivative of ( y = \frac{x+2}{x-2} ) is ( y' = \frac{-4}{(x - 2)^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.