How do you differentiate #y=ln[(x+9)^6(x+6)^2(x+5)^3]#?

1 Answer
Bill K.
Sep 18, 2017

First use properties of logarithms to expand this as a sum and then differentiate to get #dy/dx=6/(x+9)+2/(x+6)+3/(x+5)#. If you add these fractions, you can also say #dy/dx=(11x^2+139x+432)/(x^3+20x^2+129x+270)#.

Explanation:

Properties of logarithms allow us to write #y=ln[(x+9)^6(x+6)^2(x+5)^3]=ln[(x+9)^6]+ln[(x+6)^2]+ln[(x+5)^3].#

Continuing, this is also equal to

#y=6ln(x+9)+2ln(x+6)+3ln(x+5)#.

This is now easy to differentiate (and the use of the Chain Rule is trivial) to get

#dy/dx=6/(x+9)+2/(x+6)+3/(x+5)#.

Getting a common denominator gives

#dy/dx=(6(x+6)(x+5)+2(x+9)(x+5)+3(x+9)(x+6))/((x+9)(x+6)(x+5))#.

We continue to simplify:

#dy/dx=(6x^2+66x+180+2x^2+28x+90+3x^2+45x+162)/((x^2+15x+54)(x+5))#

#=(11x^2+139x+432)/(x^3+15x^2+54x+5x^2+75x+270)#

So

#dy/dx=(11x^2+139x+432)/(x^3+20x^2+129x+270)#

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