What happens to $w=\frac{x^2y^3}{z^4}$ , if $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively?

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If $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively , then $w=\dfrac{x^2y^3}{z^4}$ ............. approximately.

$1)\ \%\ 3\text{ decrease}$

$2)\ \%\ 4\text{ decrease}$

$3)\ \%\ 3\text{ increase}$

$4)\ \%\ 4\text{ increase}$

I denote the new $w$ by $w'$,

$$w'-w=\left( \frac{1.01^2.1.02^3}{1.03^4}-1\right)\times w=\left(\frac{101^2\times102^3}{103^4\times100}-1\right)\times w$$

From here should I evaluate $\dfrac{101^2\times102^3}{103^4\times100}-1$ by hand (calculator is not allowed) or there is a quicker method to get to the correct answer?

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We can use approximations here

$$w=\dfrac{x^2y^3}{z^4}$$

Taking $\ln()$ on both sides

$$\ln w=2\ln x+3\ln y-4\ln z$$

Differentiating both side gives

$$\displaystyle\frac{dw}{w}=2\frac{dx}{x}+3\frac{dy}{y}-4\frac{dz}{z}$$

Substituting values

$$\displaystyle\frac{dw}{w}=2\frac{0.01x}{x}+3\frac{0.02y}{y}-4\frac{0.03z}{z}$$

$$\displaystyle\frac{dw}{w}=0.02+0.06-0.12=-0.04$$

Therefore $$\displaystyle\frac{dw}{w}*100=-4\text{%}$$

Hence there's a decrease of $4\text{%}$ in $w$

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A rule of thumb: $1.001^n$ is about $1.00n$. More precisely, $(1+\alpha)^n\approx 1+n\alpha$ for small $\alpha$. This helps with a lot of estimation problems.

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Dividing the numerator and denominator both by $100^5$ gives $$\frac{1.01^2 . 1.02^3}{1.03^4} -1$$ Now, we use $(1+x)^n=1+nx$ for $x<<1$.

Using this, we get $(1+0.02)(1+0.06)(1-0.12)-1$ which is approximately equal to $-0.04$.

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You can use partial elasticities, also know as condition numbers:

\begin{align*} \varepsilon_w \approx & \frac{x w'_x}{w} \varepsilon_x+\frac{y w'_y}{w} \varepsilon_y + \frac{z w'_z}{w} \varepsilon_z\\ = & 2\varepsilon_x + 3 \varepsilon_y - 4 \varepsilon_z= 2 + 3\times 2-4 \times 3 = -4 \end{align*}

So, the answer is an approximate decreasing by $\text{4%}$.

This comes from Taylor's formula: \begin{align*} f(\tilde x) \approx f(x) + f'(x)(\tilde x -x) \Rightarrow \\ f(\tilde x) - f(x) \approx f'(x)(\tilde x -x) \Rightarrow \\ \frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{f'(x)(\tilde x -x)}{f(x)} \Rightarrow\\ \frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{xf'(x)}{f(x)} \frac{\tilde x -x}{x} \end{align*}