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

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.

$\frac{\frac{d}{\mathrm{dx}} \left[2\right] \cdot {\left(x + 1\right)}^{2} - 2 \cdot \frac{d}{\mathrm{dx}} \left[{\left(x + 1\right)}^{2}\right]}{{\left(x + 1\right)}^{4}}$

Before we continue, it should be noted that $\frac{d}{\mathrm{dx}} \left[2\right] = 0$, which will cause the first term in the numerator to become $0$. Also, we will have to differentiate ${\left(x + 1\right)}^{2}$, which requires the chain rule.

The chain rule states that the derivative of ${u}^{2}$ is $2 u \cdot u '$, so $\frac{d}{\mathrm{dx}} \left[{\left(x + 1\right)}^{2}\right] = 2 \left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left[x + 1\right] = 2 \left(x + 1\right) \cdot 1 = 2 x + 2$.

Going back to finding the second derivative:

$\frac{0 - 2 \left(2 x - 2\right)}{{\left(x + 1\right)}^{4}} = \frac{- 4 \left(x + 1\right)}{{\left(x + 1\right)}^{4}} = \textcolor{red}{- \frac{4}{{\left(x + 1\right)}^{3}}}$