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Let $f = f(x,y,z,w)$ where $w = w(x,y,z)$. How do you compute $\frac{\partial f}{\partial x}$? For example, say $f(x,y,z,w) = x + yze^{xw}$. Then you would get $\frac{\partial f}{\partial x} = (1 + yzwe^{xw}) + (0 + yz\frac{\partial w}{\partial x})e^{xw}$ = (take derivative with respect to x as if y,z,w is constant) + (take derivative as if x,y,z constant but w is still a function of x). So a general formula would then be \begin{equation} \frac{\partial f}{\partial x}|_{(x_0,y_0,z_0)} = \frac{\partial}{\partial x} f(x,y_0,z_0,w(x_0,y_0,z_0))|_{x = x_0} + \frac{\partial}{\partial x} f(x_0,y_0,z_0,w(x,y_0,z_0))|_{x = x_0} \end{equation}

Is this correct? If so, is there a simpler way of seeing why this formula works?

TylerMasthay
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  • The partial derivative notation $\partial/\partial x$ doesn't work well in situations like this. It doesn't specify whether $w$ is an independent variable held constant, or a dependent variable changing with $x$. – mr_e_man Sep 05 '18 at 22:47
  • You're right. My first term $\frac{\partial}{\partial x} f(x,y_0,z_0,w(x_0,y_0,z_0))|_{x = x_0}$ is treating $w$ as an independent variable held constant. My second term is treating it is a dependent variable that changes with $x$. – TylerMasthay Sep 05 '18 at 22:54
  • A better notation for that would be $$f_x(x_0,y_0,z_0,w(x_0,y_0,z_0))$$ or $$f^{(1,0,0,0)}(x_0,y_0,z_0,w(x_0,y_0,z_0))$$ – mr_e_man Sep 05 '18 at 22:56

1 Answers1

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Define a 3-variable function

$$g(x,y,z) = f(x,y,z,w(x,y,z))$$

Then its derivative, according to the chain rule, is

$$g^{(1,0,0)}(x,y,z) = f^{(1,0,0,0)}(x,y,z,w(x,y,z)) + f^{(0,0,0,1)}(x,y,z,w(x,y,z))\cdot w^{(1,0,0)}(x,y,z)$$

As another example, if

$$h(x) = f(x,5,u(x),v(x))$$

then

$$\begin{align} h'(x) = h^{(1)}(x) &= f^{(1,0,0,0)}(x,5,u(x),v(x)) \\ &+ f^{(0,0,1,0)}(x,5,u(x),v(x))\cdot u'(x) \\ &+ f^{(0,0,0,1)}(x,5,u(x),v(x))\cdot v'(x) \end{align}$$

mr_e_man
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