Question

How do you find the derivative of `ln[x^3(x+8)^7(x^2+3)^5]`?

Answer 1

`d/dx(ln[x^3(x+8)^7(x^2+3)^5])=(20x^3+104x^2+30x+72)/(x^4+8x^3+3x^2+24x)`

Explanation:

We use the combination of formulas for finding derivative of natural logarithm and derivative of product of three differentiable function UVW.

`d/dx(ln Z)=(1/Z)*d/dx(Z)` and
`d/dx(UVW)=VW*d/dx(U)+UW*d/dx(V)+UV*d/dx(W)`

Let `U=x^3` and `V=(x+8)^7` and `W=(x^2+3)^5`

We can use the formulas now

`d/dx(ln [x^3(x+8)^7(x^2+3)^5])`

`=1/(x^3(x+8)^7(x^2+3)^5)*d/dx(x^3(x+8)^7(x^2+3)^5)`

`=((x+8)^7(x^2+3)^5*d/dx(x^3)+x^3(x^2+3)^5*d/dx(x+8)^7+x^3(x+8)^7*d/dx(x^2+3)^5)/(x^3(x+8)^7(x^2+3)^5)`

`=((x+8)^7(x^2+3)^5*(3x^2)+x^3(x^2+3)^5*7(x+8)^6+x^3(x+8)^7*5(x^2+3)^4*2x)/(x^3(x+8)^7(x^2+3)^5)`

Simplify by factoring

`=(x^2(x+8)^6(x^2+3)^4)*(3(x+8)(x^2+3)+7x(x^2+3)+10x^2(x+8))/(x^3(x+8)^7(x^2+3)^5)`

`=(3(x+8)(x^2+3)+7x(x^2+3)+10x^2(x+8))/(x(x+8)(x^2+3))`

`=(3x^3+9x+24x^2+72+7x^3+21x+10x^3+80x^2)/(x(x+8)(x^2+3))`

`(20x^3+104x^2+30x+72)/(x^4+8x^3+3x^2+24x)`

God bless.... I hope the explanation is useful.

Answer 2

I use properties of logarithms to rewrite the expression and get `3/x+7/(x+8)+(10x)/(x^2+3)`.

Explanation:

Let `f(x) = ln[x^3(x+8)^7(x^2+3)^5]`

`=ln(x^3)+ln(x+8)^7+ln(x^2+3)^5`

`= 3lnx+7ln(x+8)+5ln(x^2+3)`.\

So

`f'(x) = 3/x+7/(x+8)+5/(x^2+3) d/dx(x^2+3)`

`= 3/x+7/(x+8)+(10x)/(x^2+3)`

Rewrite with a common denominator if you wish.