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

1 Answer
mason m
Nov 14, 2015

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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