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I cannot for the life of me work out the answer to this partial derivative.

$$\frac{\delta}{\delta x}\left(\frac{x^2}{(x+y)^2(x+z)^2}\right) $$

My first thought was:

  1. Split into two equations:

$$\frac{\delta}{\delta x}\left(\frac{x}{(x+y)^2}\cdot \frac{x}{(x+z)^2}\right)$$

  1. Apply chain rule to each side which gives me:

$$\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}$$ and $$ \frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3} $$

  1. I then try to use the product rule, so:

$$ \left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+z)^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+y)^2}\right\rbrack $$

However I don't end up anywhere near an answer, let alone the right answer.

Apparently the answer is:

$$ -\frac{2(x^3-xyz)}{(x+y)^3(x+z)^3} $$

Is there an easier way to do this?

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  • $\begingroup$ Post your efforts please. $\endgroup$ – the_candyman May 12 '16 at 18:10
  • $\begingroup$ @the_candyman - answer updated with efforts. $\endgroup$ – CircularRecursion May 12 '16 at 18:14
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$$\left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+{\bf{z}})^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+{\bf{y}})^2}\right\rbrack = \\ \frac{x(y-x)}{(x+y)^3(x+z)^2}+\frac{x(z-x)}{(x+z)^3(x+y)^2} = \\ \frac{x(y-x)(x+z)+x(z-x)(x+y)}{(x+y)^3(x+z)^3} = \\ \frac{x^2y -x^3 +xyz-x^2z + x^2z -x^3 + xyz -x^2y}{(x+y)^3(x+z)^3} = \\ \frac{ -2x^3 +2xyz}{(x+y)^3(x+z)^3}. $$

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