question_answer

                        Using differentials, find the approximate value of \[{{(82)}^{1/4}}\] up to three places of decimal.

Answer:

Let \[f(x)={{x}^{1/4}}.\] Then, \[f'(x)=\frac{1}{4{{x}^{3/4}}}\]             Now, \[\{f(x+\delta x)-f(x)\}=f'(x)\cdot \delta x\]             \[\Rightarrow \]   \[\{f(x+\delta x)-f(x)\}=\frac{1}{4{{x}^{3/4}}}\cdot \delta x\]     ? (i) We may write, \[82=(81+1).\]                Putting x = 81 and \[\delta x=1\] in Eq. (i), we get                         \[f\left( 81+1 \right)-f(81)=\frac{1}{4\times {{(81)}^{3/4}}}\cdot 1\]             \[\Rightarrow \]   \[f\left( 82 \right)-f(81)=\frac{1}{(4\times {{3}^{3}})}=\frac{1}{108}\] \[\Rightarrow \] \[f(82)=\left\{ f(81)+\frac{1}{108} \right\}=\left\{ {{(81)}^{1/4}}+\frac{1}{108} \right\}\]   \[=3+0.009\]             \[=3.009.\]