Question

Using differentials, find the approximate value of ${\left(\frac{17}{81}\right)}^{\frac{1}{4}}$

Answer 1

Here we will assume a function $f\left(x\right)=x^{\frac{1}{4}}$

Formula for approximation is given by

$f(x+△x)=f\left(x\right)+f^{{}^{\prime }}\left(x\right).△x$

$f^{{}^{\prime }}\left(x\right)=\frac{1}{4}{\left(x\right)}^{\frac{-3}{4}}$

Now most important step is to find $△x$ and for this we need to refer to question.

We need to choose $x$ and one important thing which we need to keep in mind is that $x$ when inserted in $f\left(x\right)$ should give a perfect outcome, here $x$ should be the nearest number to $\frac{17}{81}$ and should give a perfect fourth root.

Best number which can be seen is $\frac{16}{81}$ so from here we can conclude that $△x=\frac{1}{81}$

Plugging all the value in formula $f(x+△x)=f\left(x\right)+f^{{}^{\prime }}\left(x\right).△x$

We get ${\left(\frac{16}{81}\right)}^{\frac{1}{4}}+$ $\frac{1}{4}{\left(\frac{16}{81}\right)}^{\frac{-3}{4}}.\frac{1}{81}$

This is equal to $\frac{65}{96}$.