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

1 Answer
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)dydx=6x+9+2x+6+3x+5. If you add these fractions, you can also say dy/dx=(11x^2+139x+432)/(x^3+20x^2+129x+270)dydx=11x2+139x+432x3+20x2+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].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)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)dydx=6x+9+2x+6+3x+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))dydx=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))dydx=6x2+66x+180+2x2+28x+90+3x2+45x+162(x2+15x+54)(x+5)

=(11x^2+139x+432)/(x^3+15x^2+54x+5x^2+75x+270)=11x2+139x+432x3+15x2+54x+5x2+75x+270

So

dy/dx=(11x^2+139x+432)/(x^3+20x^2+129x+270)dydx=11x2+139x+432x3+20x2+129x+270