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How do you use the quotient rule to differentiate #(2x+1)/(x^2-1)#?

Answer 1

Luke Alvarez

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

Standard form: # (d)/dx (u/v)=(dy)/(dx)= (v (du)/(dx) - u(dv)/(dx))/v^2#

Given:#color(white)(..) y=(2x+1)/(x^2-1)#

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

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

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

But #-2x^2-2x-2= -2(x^2+x+1)#

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

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Answer 2

Samantha Adkison

To differentiate (2x+1)/(x^2-1) using the quotient rule, follow these steps:

Identify the numerator function as u(x) and the denominator function as v(x).
Apply the quotient rule formula: (u'v - uv') / v^2, where u' and v' are the derivatives of u(x) and v(x) respectively.
Find the derivatives: u'(x) = 2, v'(x) = 2x.
Substitute the derivatives and functions into the quotient rule formula.
Simplify the expression to get the derivative of the function.

The derivative of (2x+1)/(x^2-1) using the quotient rule is:

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