How to show, that a relative mistake of a special function can be estimated in a given way

Question

how to show, that if you have a function like this $$ y = f(x_1,...,x_m) := c \frac{x_1 *...*x_r}{x_{r+1},...,x_m}, \quad 1 < r \leq m,$$

the relative mistake in first order can be estimated like $$ \left | \left | \frac{\Delta y}{y} \right |\right | \leq \sum_{i=1}^{m} \left | \left | \frac{\Delta x_i}{x_i} \right | \right |.$$

Thank you very much.

Answer 1

Its $$\left | \left | \Delta y \right | \right | \leq \sum_{i=1}^{m} \left | \left | \frac{\partial f}{\partial x_i} \right | \right | \left | \left | \Delta x_i \right | \right |$$ for $i \leq r:$ $$\frac{\partial f}{\partial x_i} = c * \frac{x_1 \cdot ...\cdot x_{i-1}\cdot x_{i+1} \cdot ... \cdot x_r }{x_{r+1} \cdot ... \cdot x_m}=\frac{1}{x_i}f$$

for $r < i \leq m:$ $$\frac{\partial f}{\partial x_i} = c * \frac{x_1 \cdot ...\cdot x_r }{x_{r+1} \cdot ... \cdot (x_i)^2 \cdot x_{i+1}\cdot ... \cdot x_m}=-\frac{1}{x_i}f$$

$$\Rightarrow \left | \left | \Delta y \right | \right | \leq \sum_{i=1}^{r} \left | \left | \frac{1}{x_i} f\right | \right | \left | \left | \Delta x_i \right | \right | + \sum_{i>r}^{m} \left | \left | - \frac{1}{x_i} f\right | \right | \left | \left | \Delta x_i \right | \right | = \sum_{i=1}^{m} \left | \left | \frac{1}{x_i} f\right | \right | \left | \left | \Delta x_i \right | \right | = \left | \left | f \right | \right | \sum_{i=1}^{m} \left | \left | \frac{\Delta x_i}{x_i} \right | \right |$$

$$\Rightarrow \frac{\left | \left | \Delta y \right | \right |}{\left | \left | f \right | \right |} \leq \sum_{i=1}^{m} \left | \left | \frac{\Delta x_i}{x_i} \right | \right |$$

$$\Rightarrow \frac{\left | \left | \Delta y \right | \right |}{\left | \left | y \right | \right |} \leq \sum_{i=1}^{m} \left | \left | \frac{\Delta x_i}{x_i} \right | \right |$$