Question

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))`