Question

Using differentials, find the approximate value of ${\left(33\right)}^{\frac{1}{5}}$

Answer 1

Here we will assume a function $f\left(x\right)=x^{\frac{1}{5}}$

Formula for approximation is given by

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

$f^{{}^{\prime }}\left(x\right)=\frac{1}{5}{\left(x\right)}^{\frac{-4}{5}}$

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 nearest number to $33$ and should give a perfect fifth root.

Best number which can be seen is $32$ so from here we can conclude that $△x=1$

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

We get ${\left(32\right)}^{\frac{1}{5}}+$ $\frac{1}{5}{\left(32\right)}^{\frac{-4}{5}}.1$

This is equal to $\frac{161}{80}$