How do you find the first and second derivatives of #f(x)=(x-1)/(x+1)# using the quotient rule?

Answer 1
Remember that the quotient rule says: #d/(dx)[(f(x))/(g(x))]=(f'(x)g(x)-f(x)g'(x))/((g(x))^2#
Finding the first derivative: #(d/(dx)[x-1]*(x+1)-(x-1)*d/(dx)[x+1])/((x+1)^2)=(1⋅(x+1)-(x-1)⋅1)/((x+1)^2)=((x+1)-(x-1))/((x+1)^2)=color(blue)(2/((x+1)^2))#

That is the first derivative. To find the second derivative, use the quotient rule on the first derivative.

#(d/(dx)[2]*(x+1)^2-2*d/(dx)[(x+1)^2])/((x+1)^4)#
Before we continue, it should be noted that #d/(dx)[2]=0#, which will cause the first term in the numerator to become #0#. Also, we will have to differentiate #(x+1)^2#, which requires the chain rule.
The chain rule states that the derivative of #u^2# is #2u*u'#, so #d/(dx)[(x+1)^2]=2(x+1)*d/(dx)[x+1]=2(x+1)*1=2x+2#.

Going back to finding the second derivative:

#(0-2(2x-2))/((x+1)^4)=(-4(x+1))/((x+1)^4)=color(red)(-4/((x+1)^3))#
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Answer 2

To find the first and second derivatives of ( f(x) = \frac{x - 1}{x + 1} ) using the quotient rule:

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

Second Derivative: [ f''(x) = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} - \frac{2(x - 1)((x + 1)(1) - (x - 1)(1))}{(x + 1)^3} ]

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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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