Question

Find the first partial derivates of \(W = \ln \left( {x + 2y + 3z} \right)\).

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

Step (1):- Finding partial derivative with respect to x.

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

Step (2):- Finding partial derivative with respect to y.

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

Step (3):- finding 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, first partial derivative are

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