How do we differentiate #ln((x^2)/((x + 3)(x^2 - 1)))#?

Question

354 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-08T15:04:53

#d/dxln((x^2)/((x + 3)(x^2 - 1))) = 2/x - 1/(x + 3) - 1/(x + 1) - 1/(x - 1)#

We use the properties of logarithms to simplify first.

Call the function #y#. We use the rules #ln(a/n) = lna - ln n# and #ln(an)= lna + lnn#.

#y = ln(x^2) - ln(x + 3) - ln(x^2 - 1)#

Note that #x^2 - 1 = (x + 1)(x - 1)#.

#y = ln(x^2) - ln(x +3) - ln((x + 1)(x -1))#

#y = ln(x^2) - ln(x + 3) - (ln(x + 1) + ln(x - 1))#

#y = ln(x^2) - ln(x + 3) - ln(x + 1) - ln(x - 1)#

Now use #ln(a^n) = nlna#.

#y = 2lnx - ln(x + 3) - ln(x + 1) - ln(x -1)#

We know that #d/dx(lnx) = 1/x#.

#y' =2/x - 1/(x + 3) - 1/(x + 1) - 1/(x - 1)#

This is our answer.

Hopefully this helps!