here's the following:
$$u = f(x,xy,xyz)$$
I am pretty new to partial derivatives, so do not judge much, but whats big difference (in the solving), between for example
$$u = x^4 + y^4 -4x^2y^2$$
where we take it as $\frac{\partial{u}}{\partial{x}}$ then as $\frac{\partial{u}}{\partial{y}}$ and finally as $\frac{\partial{^2u}}{\partial{x}\partial{y}}$
and the second one?
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$\begingroup$I will answer the question I think you are asking.
Suppose $$ u = f(x,xy,xyz) . $$
Then to find $\partial u / \partial x$ you must use the chain rule: $$ \frac{\partial u}{\partial x} = f_1(x,xy,xyz) + yf_2(x,xy,xyz) + yzf_3(x,xy,xyz) $$ where I've written $f_1$ for the partial derivative of $f$ with respect to its first variable, and so on.
If I'm correct you should be able to do the rest. It is ugly but straightforward using the sum, product and chain rules: $$ \begin{align} \frac{\partial u}{\partial y \partial x} & = \frac{\partial}{\partial y}f_1(x,xy,xyz) + \frac{\partial}{\partial y} yf_2(x,xy,xyz) + \frac{\partial}{\partial y} yzf_3(x,xy,xyz) \\ &= xf_{21}(x,xy,xyz) + xzf_{31}(x,xy,xyz) \\ & \quad + f_2(x,xy,xyz) + y\left( xf_{22}(x,xy,xyz) + xzf_{32}(x,xy,xyz) \right)\\ & \quad + \text{and so on} \end{align} $$
$\endgroup$ 2 $\begingroup$I'm not sure exactly what you're asking, but here is how to compute the partial derivatives:
$$u = x^4+y^4-4x^2y^2$$
When taking the partial derivative with respect to a variable $x$, you differentiate as normal whilst treating other variables as constants.
$$\frac{\partial u}{\partial x} = 4x^3-8y^2x$$
$$\frac{\partial ^2u}{\partial y \partial x} = -16yx$$
$$\frac{\partial u}{\partial y} = 4y^3-8x^2y$$
$$\frac{\partial^2u}{\partial x\partial y} = -16yx = \frac{\partial ^2u}{\partial y \partial x}$$
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