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

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Answer 1 · 2015-12-30T03:47:48

$f'(x)=(2x+2)/(x^2+2x-3)$

$f''(x)=-(2(x^2+2x+5))/(x^2+2x-3)^2$

Use the chain rule -- if $f(x)=lnu$ , then $f'(x)=1/u*u'$ .

Thus,

$f'(x)=1/(x^2+2x-3)*d/dx(x^2+2x-3)$

$=>(2x+2)/(x^2+2x-3)$

Use the quotient rule to find the second derivative.

$f''(x)=((x^2+2x-3)d/dx(2x+2)-(2x+2)d/dx(x^2+2x-3))/(x^2+2x-3)^2$

$=>(2(x^2+2x-3)-(2x+2)^2)/(x^2+2x-3)^2$

Continued simplification yields:

$f''(x)=-(2(x^2+2x+5))/(x^2+2x-3)^2$

What are the first and second derivatives of f(x)=ln(x^2+2x-3) f(x)=ln(x2+2x−3)f(x)=ln(x^2+2x-3) ?