Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.

$(15{)}^{\frac{1}{4}}$

Answer 1

Consider $y=x^{\frac{1}{4}}$.

Let $x=16$ and $\Delta x=-1$.

Then,

$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(15{)}^{\frac{1}{4}}-(16{)}^{\frac{1}{4}}=(15{)}^{\frac{1}{4}}-2$

$\Rightarrow (15{)}^{\frac{1}{4}}=2+\Delta y$

Now, $dy$ is approximately equal to $\Delta$y and is given by,

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{4(x{)}^{\frac{3}{4}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{4}}$]

$=\frac{1}{4(16{)}^{\frac{3}{4}}}(-1)=\frac{-1}{4\times 8}=\frac{-1}{32}=-0.03125$

Hence, the approximate value of $(15{)}^{\frac{1}{4}}$ is $2+(-0.03125)=1.96875\approx 1.968$.