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

Answer 1

Toolbox:

• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$

Step 1:

$\big(\large\frac{17}{18}\big)^{\Large\frac{1}{4}}=\frac{(17)^{\Large\frac{1}{4}}}{(81)^{\Large\frac{1}{4}}}$

$\qquad\qquad=\large\frac{(17)^{\Large\frac{1}{4}}}{3}$

Let $f(x)=x^{\Large\frac{1}{4}}$

$\Rightarrow x=16$, $\Delta x=1$

$f(x+\Delta x)=(17)^{\Large\frac{1}{4}}$

$\Delta y=f(x+\Delta x)-f(x)$

$f(x+\Delta x)=f(x)+\Delta y$

Step 2:

$dy$ is approximately equal to $\Delta y$

$dy=\large\frac{dy}{dx}$$\times\Delta x$

$f(x)=x^{\Large\frac{1}{4}}$

Differentiating with respect to x we get

$f'(x)=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$

$dy=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$$\Delta x$

$\quad=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$$.1$

$\Rightarrow \large\frac{1}{4(16)^{\Large\frac{3}{4}}}$$\times 1$

$\Rightarrow \large\frac{1}{4(8)}$$\times 1$

$\Rightarrow \large\frac{1}{32}$

Step 3:

$(17)^{\Large\frac{1}{4}}=2+\large\frac{1}{32}$$=2.03125$

$\big(\large\frac{17}{81}\big)^{\Large\frac{1}{4}}=\frac{(17)^{\Large\frac{1}{4}}}{3}$

$\qquad\;\;\;\;\;=\large\frac{2.03125}{3}$$=0.677083$

$\qquad\;\;\;\;\;=0.677$(Approx)