Question
Using differentials, find the approximate values of the following:
$(80)^{\frac{1}{4}}$
$(80)^{\frac{1}{4}}$
✓
Answer
Let $\text{x}=81,\\\text{ x}+ \triangle\text{x}=80$
$\triangle\text{x}=80- 81$
$=-1$
Let $\text{y}=\text{x}^ {\frac{1}{4}}$
$\frac{\text{dy}} {\text{dx}}=\frac{1}{4(81)^{\frac{3} {4}}}$
$=\frac{1}{108}$
$0.00926$
$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=81}\times(\triangle\text{x})$
$=0.00926)(-1)$
$=-0.00926$
$(80)^{\frac{1}{4}}= \text{y}+\triangle\text{y}$
$=\text{x}^{\frac{1} {4}}-0.00926$
$=(81)^{\frac{1}{4}}- 0.00926$
$3-0.0026$
$2.9974$
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