Using differentials, find the approximate value of each of the following.

(a) Consider $y=x^{\frac{1}{4}}$. Let $x=\frac{16}{81}$ and $\Delta x=\frac{1}{81}$.

Then, $\Delta y=(x+\Delta x)^{\frac{1}{4}}-x^{\frac{1}{4}}$

$=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\left(\frac{16}{81}\right)^{\frac{1}{4}}$

$=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\frac{2}{3}$

$\therefore\left(\frac{17}{81}\right)^{\frac{1}{4}}=\frac{2}{3}+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$\begin{aligned} d y &=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left(\text { as } y=x^{\frac{1}{4}}\right) \\ &=\frac{1}{4\left(\frac{16}{81}\right)^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4 \times 8} \times \frac{1}{81}=\frac{1}{32 \times 3}=\frac{1}{96}=0.010 \end{aligned}$

Hence, the approximate value of $\left(\frac{17}{81}\right)^{\frac{1}{4}}$ is $\frac{2}{3}+0.010=0.667+0.010$

$=0.677$

(b) Consider $y=x^{\frac{1}{5}}$. Let $x=32$ and $\Delta x=1$.

Then, $\Delta y=(x+\Delta x)^{-\frac{1}{5}}-x^{-\frac{1}{5}}=(33)^{-\frac{1}{5}}-(32)^{-\frac{1}{5}}=(33)^{-\frac{1}{5}}-\frac{1}{2}$

$\therefore(33)^{-\frac{1}{5}}=\frac{1}{2}+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right)(\Delta x)=\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x) \quad\left(\right.$ as $\left.y=x^{\frac{1}{5}}\right)$

$=-\frac{1}{5(2)^{6}}(1)=-\frac{1}{320}=-0.003$

Hence, the approximate value of $(33)^{-\frac{1}{5}}$ is $\frac{1}{2}+(-0.003)$

= 0.5 − 0.003 = 0.497.