The value of $w$ also has a max error of $p\%$

Question

Suppose $\frac{1}{w}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ where each variable $x,y,z$ can be measured with a max error of $p\%$

Prove that the calculated value of $w$ also has a max error of $p\%$

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I guess I need to take its derivative i.e $-\frac{1}{w^2}dw=-\frac{1}{x^2}dx-\frac{1}{y^2}dy-\frac{1}{z^2}dz$

After there, what do I need to do?

I know this question maybe seem nonsense. But please help me to solve. Thanks!:)

Answer 1

From this line

$$-\frac{1}{w^2}dw=-\frac{1}{x^2}dx-\frac{1}{y^2}dy-\frac{1}{z^2}dz$$

The percentage error of $w$ is

$$p_w=100\frac{dw}{w}\%$$

and likewise for the other variables, therefore

$$\frac{1}{w}p_w=\frac{1}{x}p_x+\frac{1}{y}p_y+\frac{1}{z}p_z$$

So assuming a maximum error of $p\%$ on the independent variables we have that

$$\frac{1}{w}p_w\leq\frac{1}{x}p+\frac{1}{y}p+\frac{1}{z}p=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)p=\frac{1}{w}p$$