如何求解包含可导函数的多元函数的一阶偏导数？

Let's walk through exactly how to compute the first-order partial derivatives for your example function $F(x,y)=x^2 y h(x^2+y2)$—where $h$ is differentiable—and break down the general rules you can apply to similar problems.

1. 对x的偏导数 $\partial F/\partial x$ #

When taking the partial derivative with respect to $x$, we treat $y$ as a constant. This function is a product of two terms: $u = x^2 y$ and $v = h(x^2 + y^2)$, so we'll use the product rule ($\partial(uv)/\partial x = (\partial u/\partial x)v + u(\partial v/\partial x)$) plus the chain rule for the nested $h$ function:

• Compute $\partial u/\partial x$: Since $y$ is constant, this is $2xy$.
• Compute $\partial v/\partial x$: Apply the chain rule—first take the derivative of $h$ with respect to its argument, then multiply by the partial derivative of the argument $x^2 + y^2$ with respect to $x$. That gives $h'(x^2 + y^2) \cdot 2x$.
• Combine using the product rule:

  ∂F/∂x = 2xy * h(x²+y²) + x²y * [h'(x²+y²) * 2x]
  

  Simplify it to:

  ∂F/∂x = 2xy h(x²+y²) + 2x³y h'(x²+y²)
  

2. 对y的偏导数 $\partial F/\partial y$ #

Now treat $x$ as a constant and repeat the process:

• Again, use the product rule with $u = x^2 y$ and $v = h(x^2 + y^2)$:
  • $\partial u/\partial y$ is $x^2$ (since $x$ is constant).
  • $\partial v/\partial y$ uses the chain rule: derivative of $h$ times partial derivative of $x^2 + y^2$ with respect to $y$, so $h'(x^2 + y^2) \cdot 2y$.
• Combine the terms:

  ∂F/∂y = x² * h(x²+y²) + x²y * [h'(x²+y²) * 2y]
  

  Simplify to:

  ∂F/∂y = x² h(x²+y²) + 2x²y² h'(x²+y²)
  

通用求解步骤总结 #

• Spot composite functions: Anytime you have a function like $h(g(x,y))$, remember the chain rule works for partial derivatives too—differentiate the outer function first, then multiply by the partial derivative of the inner function with respect to your target variable.
• Apply product rule for products: If your function is a product of multiple terms (like $x^2 y$ times $h(...) here$), split it into parts, differentiate each part separately, and combine using the product rule.
• Treat other variables as constants: When taking a partial derivative with respect to $x$, $y$ (and any other variables) don't change—so their derivatives are zero, and you can factor them out like constants during differentiation.

内容的提问来源于stack exchange，提问作者Edward B