# What are the first and second derivatives of f(x)=ln(x^2+2x-3) ?

Dec 30, 2015

$f ' \left(x\right) = \frac{2 x + 2}{{x}^{2} + 2 x - 3}$

$f ' ' \left(x\right) = - \frac{2 \left({x}^{2} + 2 x + 5\right)}{{x}^{2} + 2 x - 3} ^ 2$

#### Explanation:

Use the chain rule -- if $f \left(x\right) = \ln u$, then $f ' \left(x\right) = \frac{1}{u} \cdot u '$.

Thus,

$f ' \left(x\right) = \frac{1}{{x}^{2} + 2 x - 3} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x - 3\right)$

$\implies \frac{2 x + 2}{{x}^{2} + 2 x - 3}$

Use the quotient rule to find the second derivative.

$f ' ' \left(x\right) = \frac{\left({x}^{2} + 2 x - 3\right) \frac{d}{\mathrm{dx}} \left(2 x + 2\right) - \left(2 x + 2\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x - 3\right)}{{x}^{2} + 2 x - 3} ^ 2$

$\implies \frac{2 \left({x}^{2} + 2 x - 3\right) - {\left(2 x + 2\right)}^{2}}{{x}^{2} + 2 x - 3} ^ 2$

Continued simplification yields:

$f ' ' \left(x\right) = - \frac{2 \left({x}^{2} + 2 x + 5\right)}{{x}^{2} + 2 x - 3} ^ 2$