What are the first and second derivatives of #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#