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I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.

1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.

2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.

3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.

Why 2) and 3) are not correct?

jupiter_jazz
  • 517

2 Answers2

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Note: $$\frac{\partial}{\partial x}\left( \left(\sqrt{x^2+y^2}\right)^{-1}\right)=(-1)\left(\sqrt{x^2+y^2}\right)^{-2}\frac{\partial}{\partial x}\sqrt{x^2+y^2}$$ $$\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)=\frac{0-1\cdot \frac{\partial}{\partial x}\sqrt{x^2+y^2}}{\left(\sqrt{x^2+y^2}\right)^2}$$

farruhota
  • 31,482
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In your calculations you should use these 3 formulas:

  • $\frac{\partial}{\partial x}\frac{1}{a(x,y)} = \frac{-1}{a^2(x,y)} \cdot \frac{\partial}{\partial x} a(x,y)$
  • $\frac{\partial}{\partial x}\sqrt{b(x,y)} = \frac{1}{2\sqrt{b(x,y)}} \cdot \frac{\partial}{\partial x} b(x,y)$
  • $\frac{\partial}{\partial x}(x^2+c(y)) = 2x$

In point 2:

  • $c(y)=y^2$
  • $a(x,y) = x^2 + c(y)$
  • $b(x,y) = \frac{1}{a(x,y)}$
  • Compute $\frac{\partial}{\partial x} \sqrt{b(x,y)}$

In point 3 :

  • $c(y)=y^2$
  • $b(x,y) = x^2+c(y) $
  • $a(x,y) = \sqrt{b(x,y)} $
  • Compute $\frac{\partial}{\partial x} \frac{1}{a(x,y)}$
Jaroslaw Matlak
  • 4,895