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# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(vi)\;(15)^{\Large\frac{1}{4}}$

This is sixth part of multipart q1

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:
Let $y=x^{\Large\frac{1}{4}}$
Also let $x=16$
$\Delta x=-1$
So that $x+\Delta x=15$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{4}}-x^{\Large\frac{1}{4}}$
$\quad\;\;=(15)^{\Large\frac{1}{4}}-(16)^{\Large\frac{1}{4}}$
$\quad\;\;=(15)^{\Large\frac{1}{4}}-2$
$\Delta y=(15)^{\Large\frac{1}{4}}-2$
$\Delta y+2=(15)^{\Large\frac{1}{4}}$
$(15)^{\Large\frac{1}{4}}=\Delta y+2$
$\large\frac{dy}{dx}=\large\frac{1}{4}x^{-\Large\frac{3}{4}}$
$\quad=\large\frac{1}{4x^{\Large\frac{3}{4}}}$
Step 2:
Now $\Delta y$ is approximately equal to $dy$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x \quad=\large\frac{1}{4x^{\Large\frac{3}{4}}}$$\Delta x$
$\quad=\large\frac{1}{4(16)^{\Large\frac{3}{4}}}$$\times (-1)$
$\quad=\large\frac{-1}{4\times 8}=\frac{-1}{32}$
$\quad=-0.03125$
Step 3:
Approximate value of $\Delta y=dy=0.03125$
Hence from equation(1) we have
$(15)^{\Large\frac{1}{4}}=2-0.03125$
$\qquad\;\;=1.969$