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How do you differentiate #f(x)=(2x+1)/(2x-1)#?

Answer 1

Evelyn Agard

#d/dx ((2x+1)/(2x-1)) =-4/(2x-1)^2#

Using the quotient rule:

#d/dx ((2x+1)/(2x-1)) = ( (2x-1) d/dx (2x+1) - (2x+1)d/dx(2x-1))/(2x-1)^2#

#d/dx ((2x+1)/(2x-1)) = ( 2(2x-1) - 2(2x+1))/(2x-1)^2#

#d/dx ((2x+1)/(2x-1)) = ( 4x-2 - 4x-2)/(2x-1)^2#

#d/dx ((2x+1)/(2x-1)) =-4/(2x-1)^2#

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Answer 2

Cameron Alexander

To differentiate ( f(x) = \frac{{2x+1}}{{2x-1}} ), you can use the quotient rule of differentiation. The quotient rule states that if you have a function ( u(x) ) divided by ( v(x) ), then the derivative of ( \frac{{u(x)}}{{v(x)}} ) with respect to ( x ) is given by ( \frac{{u'(x)v(x) - u(x)v'(x)}}{{[v(x)]^2}} ).

Applying the quotient rule to the function ( f(x) = \frac{{2x+1}}{{2x-1}} ), we get:

[ f'(x) = \frac{{(2)(2x-1) - (2x+1)(2)}}{{(2x-1)^2}} ]

[ f'(x) = \frac{{4x - 2 - 4x - 2}}{{(2x-1)^2}} ]

[ f'(x) = \frac{{-4}}{{(2x-1)^2}} ]

So, the derivative of ( f(x) ) with respect to ( x ) is ( f'(x) = \frac{{-4}}{{(2x-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.