# Given y=lnx*ln(x+1)*ln(x+2), what is dy/dx?

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### 1 Answer

To find out the derivative of the given function, we'll have to use product rule:

(u*v*w) = u'*v*w + u*v'*w + u*v*w'

Let u(x) = ln x => u' = 1/x

Let v(x) = ln(x+1)=> v'(x)= (x+1)'/(x+1) => v'(x) = 1/(x+1)

Let w(x) = ln(x+2) = >w'(x) = (x+2')/(x+2) => w'(x) = 1/(x+2)

dy/dx = [ln(x+1)]*[ln(x+2)]/x + [ln(x)]*[ln(x+2)]/(x+1) + [ln(x)]*[ln(x+1)]/(x+2)

dy/dx = {(x+1)(x+2)[ln(x+1)]*[ln(x+2)] + x(x+2)[ln(x)]*[ln(x+2)] + x(x+1)[ln(x)]*[ln(x+1)]}/x(x+1)(x+2)

We'll use the power rule of logarithms:

x*ln x = ln `x^x`

(x+1)ln(x+1) = ln `(x+1)^(x+1)`

(x+1)ln(x+1) = ln `(x+2)^(x+2)`

**dy/dx = {[ln`(x+1)^(x+1)`][ln `(x+2)^(x+2)`]+ [ln`x^(x)`][ln `(x+2)^(x+2)`] + [ln `x^x`][ln `(x+1)^(x+1)`]}/x(x+1)(x+2)**