Assume function $u=u(x)$ is defined with that system of equations: $ \begin{cases} u=f(x,y,z)\\ g(x,y,z)=0\\ h(x,y,z)=0 \end{cases} $ How can i find $du/dx$?
Please help, i don't know how to start solving this..
Assume function $u=u(x)$ is defined with that system of equations: $ \begin{cases} u=f(x,y,z)\\ g(x,y,z)=0\\ h(x,y,z)=0 \end{cases} $ How can i find $du/dx$?
Please help, i don't know how to start solving this..
Start by using the chain rule. \begin{equation} u(x) = f(x,y(x),z(x)) \end{equation} so \begin{align*} u'(x) &= \frac{\partial f(x,y(x),z(x))}{\partial x} + \frac{\partial f(x,y(x),z(x))}{\partial y} y'(x) + \frac{\partial f(x,y(x),z(x))}{\partial z} z'(x). \end{align*}
Implicit differentiation allows us to find formulas for $y'(x)$ and $z'(x)$. \begin{equation} g(x,y(x),z(x)) = 0 \end{equation} so \begin{equation} \frac{\partial g(x,y(x),z(x))}{\partial x} + \frac{\partial g(x,y(x),z(x))}{\partial y}y'(x) + \frac{\partial g(x,y(x),z(x))}{\partial z}z'(x) = 0 \end{equation} and similarly \begin{equation} \frac{\partial h(x,y(x),z(x))}{\partial x} + \frac{\partial h(x,y(x),z(x))}{\partial y}y'(x) + \frac{\partial h(x,y(x),z(x))}{\partial z}z'(x) = 0. \end{equation}
These equations can be used to solve for $y'(x)$ and $z'(x)$.