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?
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 $w$