Question

Using differential, find the approximate value of the following:
$(80{)}^{1/4}$

Answer 1

Consider the function $y=f\left(x\right)=(x{)}^{\frac{1}{4}}$

Let $x=81$

$x+\Delta x=80$

Then,

$\Delta x=-1$

For $x=81,$

$\Rightarrow$ $y=(81{)}^{\frac{1}{4}}=3$

Let $dx=\Delta x=-1$

Now, $y=(x{)}^{\frac{1}{4}}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{4(x{)}^{\frac{3}{4}}}$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{x=81}=\frac{1}{108}$

$\therefore$ $\Delta y=\frac{dy}{dx}dx=\frac{1}{108}\times (-1)=-0.009259$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{x=81}=\frac{1}{108}$

$\therefore$ $\Delta y=-0.009259$

$\therefore$ $(80{)}^{\frac{1}{4}}=y+\Delta y$

$=3-0.009259$

$=2.99074$