Question

find the first partial derivates of w= ln(x+2y+3z).

Short answer

\(\frac{{\partial W}}{{\partial x}} = \frac{1}{{x + 2y + 3z}},\frac{{\partial W}}{{\partial y}} = \frac{2}{{x + 2y + 3z}},\frac{{\partial W}}{{\partial z}} = \frac{3}{{x + 2y + 3z}}\)

Step-by-step solution

Step1:- finding partial derivative with the respect to x.

w= ln(x+2y+3z)

\(\begin{aligned}{l}\frac{{\partial W}}{{\partial x}} &= \frac{\partial }{{\partial x}}\ln \left( {x + 2y + 3z} \right)\\\frac{{\partial W}}{{\partial x}} &= \frac{1}{{x + 2y + 3z}}\end{aligned}\)

step2:- finding the partial derivative with the respect to y.

\(\begin{aligned}{l}W &= \ln \left( {x + 2y + 3z} \right)\\\frac{{\partial W}}{{\partial y}} &= \frac{\partial }{{\partial y}}\ln \left( {x + 2y + 3z} \right)\\\frac{{\partial W}}{{\partial y}} &= \frac{2}{{x + 2y + 3z}}\end{aligned}\)

step3:- finding the partial derivative with respect to z.

\(\begin{aligned}{l}W &= \ln \left( {x + 2y + 3z} \right)\\\frac{{\partial W}}{{\partial z}} &= \frac{\partial }{{\partial z}}\ln \left( {x + 2y + 3z} \right)\\\frac{{\partial W}}{{\partial z}} &= \frac{3}{{x + 2y + 3z}}\end{aligned}\)

Hence , the first partial derivates are

\(\frac{{\partial W}}{{\partial x}} = \frac{1}{{x + 2y + 3z}},\frac{{\partial W}}{{\partial y}} = \frac{2}{{x + 2y + 3z}},\frac{{\partial W}}{{\partial z}} = \frac{3}{{x + 2y + 3z}}\).