# finding $du/dx$ if $u=u(x)$ is defined with system of equations

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 $du/dx$?

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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

Start by using the chain rule. $$u(x) = f(x,y(x),z(x))$$ 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 $z'(x)$. $$g(x,y(x),z(x)) = 0$$ so $$\frac{\partial g(x,y(x),z(x))}{\partial x} + \frac{\partial g(x,y(x),z(x))}{\partial y}y'(x) + \frac{\partial g(x,y(x),z(x))}{\partial z}z'(x) = 0$$ and similarly $$\frac{\partial h(x,y(x),z(x))}{\partial x} + \frac{\partial h(x,y(x),z(x))}{\partial y}y'(x) + \frac{\partial h(x,y(x),z(x))}{\partial z}z'(x) = 0.$$
These equations can be used to solve for $y'(x)$ and $z'(x)$.