Question

Use differentials to find an approximate value for the given number $y=\sqrt[3]{31.02}+\sqrt[4]{41.02}$.

Answer 1

Let $y=f\left(x\right)=x^{\frac{1}{3}}$
Take $x=1,dx=\Delta x=0.02$
$dy=\frac{1}{3}{x}^{\frac{-2}{3}}.dx=\frac{1}{3}\times \left(1\right)\left(0.02\right)=0.0066$
$f(x+\Delta x)=y+dx=f\left(1\right)+0.0066$
$f(1+0.02)=1+0.0066$
$(1.02{)}^{\frac{1}{3}}=1.0066$
Again, let $y=f\left(x\right)=x^{\frac{1}{4}}$
Here, $x=1,dx=\Delta x=0.02$
$dx=\frac{1}{4}{x}^{\frac{3}{4}}.dx=\frac{1}{4}\times \left(1\right)\left(0.02\right)=0.005$
$f(x+\Delta x)=y+dx=f\left(1\right)=0.005=1.005$
$(1.02{)}^{\frac{1}{4}}\simeq 1.005$
$(1.02{)}^{\frac{1}{3}}+(1.02{)}^{\frac{1}{4}}\simeq 1.0066+1.005=2.0116$