Question

Compute the jacobian matrices using the chain rule for z=u2+v2 .
where u=2x+7 , v=3x+y+7 .

Answer 1

Answer on Question # 41380, Math, Integral Calculus

Compute the jacobian matrices using the chain rule for $z = u2 + v2$. where $u = 2x + 7$, $v = 3x + y + 7$.

Solution:

We have function $z(u,v)$ depends on 2 variables $u$ and $v$. So the Jacobian matrix is the next:

Jacobian matrix of composition is equal to product of Jacobian matrices. $J_{\psi \circ \varphi}(x) = J_{\psi}(\varphi (x))J_{\varphi}(x)$

$$\varphi (u, v) = u ^ {2} + v ^ {2}$$ $$\vec {y} (x, y): \left( \begin{array}{c} u = 2 x + 7 \\ v = 3 x + y + 7 \end{array} \right)$$ $$J _ {\varphi} = \left(\frac {\partial \varphi}{\partial u} \frac {\partial \varphi}{\partial v}\right) = (2 u 2 v) = (4 x + 1 4 6 x + 2 y + 1 4)$$ $$J _ {\psi} = \dots \left( \begin{array}{c c} \frac {\partial u}{\partial x} & \frac {\partial u}{\partial y} \\ \frac {\partial v}{\partial x} & \frac {\partial v}{\partial y} \end{array} \right) = \left( \begin{array}{c c} 2 & 0 \\ 3 & 1 \end{array} \right)$$ $$\begin{array}{l} J = J _ {\varphi} * J _ {\psi} = (4 x + 1 4 \quad 6 x + 2 y + 1 4) \left( \begin{array}{c c} 2 & 0 \\ 3 & 1 \end{array} \right) \\ = (2 4 x + 6 y + 7 0 \quad 6 x + 2 y + 1 4) \\ \end{array}$$

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