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Whitepunk [10]
2 years ago
11

If z=(x+y)ey and x=u2+v2 and y=u2−v2, find the following partial derivatives using the chain rule. Enter your answers as functio

ns of u and v.
∂z/∂u= ______
∂z/∂v= ______
Mathematics
1 answer:
velikii [3]2 years ago
4 0

Answer:

Step-by-step explanation:

Given the functions  z=(x+y)e^y\\ and x=u²+v² and y=u²−v²

Using the composite derivative formula;

∂z/∂u= ∂z/∂x*∂x/∂u+∂z/∂y*∂y/∂u

∂z/∂u = ye^y*2u + [(x+y)e^y+xe^y]*2u

∂z/∂u =ye^y*2u + 2u[xe^y+ye^y+xe^y]

∂z/∂u = ye^y*2u + 2u[2xe^y+ye^y]

<em>∂z/∂u = 2u[u²−v²]</em>e^{u^2-v^2}<em>+ 2u[2(u²+v²)</em>e^{u^2-v^2}<em>+y</em>e^{u^2-v^2}<em>]]</em>

∂z/∂v= ∂z/∂x*∂x/∂v+∂z/∂y*∂y/∂v

∂z/∂v = ye^y*2v + [(x+y)e^y+xe^y]*-2v

∂z/∂v =ye^y*2v -2v[xe^y+ye^y+xe^y]

∂z/∂v = ye^y*2v -2v[2xe^y+ye^y]

<em>∂z/∂v = 2v[u²−v²]</em>e^{u^2-v^2}<em>-2v[2(u²+v²)</em>e^{u^2-v^2}<em>+y</em>e^{u^2-v^2}<em>]</em>

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So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

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The significance level assumed on this case is \alpha=0.05

The standard error for the slope is given by this formula:

SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}

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