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kodGreya [7K]
3 years ago
8

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

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
7 0

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)

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