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

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