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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..

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You take the expression of $f$ and calculate the derivative as if $y$ and $z$ were constants. – xavierm02 Nov 18 '12 at 10:51
For what's seen, and if not more info is given: $$\frac{du}{dx}=\frac{\partial f}{\partial x}$$ – DonAntonio Nov 18 '12 at 10:52
up vote 3 down vote accepted

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 $z'(x)$.

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