Question

Use the chain rule calculate dw/dr , dw/ds and dw/dt

Answer 1

Answer:

$\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$, $\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$, $\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$

Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be $w = \ln (x+2\cdot y + 3\cdot z)$, $x = r^{2}+t^{2}$, $y = s^{2}-t^{2}$ and $z = r^{2}+s^{2}$:

$\frac{dw}{dr} = \frac{\frac{dx}{dr}+2\cdot \frac{dy}{dr} +3\cdot \frac{dz}{dr} }{x+2\cdot y + 3\cdot z}$

$\frac{dx}{dr} = 2\cdot r$

$\frac{dy}{dr} = 0$

$\frac{dz}{dr} = 2\cdot r$

$\frac{dw}{dr} = \frac{8\cdot r}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}$

$\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$ (1)

$\frac{dw}{ds} = \frac{\frac{dx}{ds}+2\cdot \frac{dy}{ds} +3\cdot \frac{dz}{ds} }{x+2\cdot y + 3\cdot z}$

$\frac{dx}{ds} = 0$

$\frac{dy}{ds} = 2\cdot s$

$\frac{dz}{ds} = 2\cdot s$

$\frac{dw}{ds} = \frac{10\cdot s}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}$

$\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$ (2)

$\frac{dw}{dt} = \frac{\frac{dx}{dt}+2\cdot \frac{dy}{dt} +3\cdot \frac{dz}{dt} }{x+2\cdot y + 3\cdot z}$

$\frac{dx}{dt} = 2\cdot t$

$\frac{dy}{dt} = -2\cdot t$

$\frac{dz}{dt} = 0$

$\frac{dw}{dt} = -\frac{2\cdot t}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}$

$\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$ (3)