# Use differentials to find an approximate value for the given numbers $\sqrt{36.1}$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• Let $y=f(x)$ be a differentiable function then the quantities $dx$ and $dy$ are called differentials.The differential $dx$ is an independent variable.
• The differential $dy$ is then defined by $dy=f'(x)dx(dx \approx \Delta x)$
• Also $f(x+\Delta x)-f(x)=\Delta y =dy$ from which $f(x+\Delta x)$can be evaluated
Given $\sqrt {36.1}$
Step 1:
Let $y=f(x)=\sqrt x =x^{\large\frac{1}{2}}$
$dy=\large\frac{1}{2}$$x^{\large\frac{-1}{2}}$$dx$
Now $f(36)=\sqrt {36}=6$
Step 2:
We take $x=36$ and $\Delta x \approx \alpha x =0.1$
This gives
$dy=\large\frac{1}{2} $$(36)^{\large\frac{-1}{2}}$$(1)=\large\frac{1}{12}$$=0.00833$
$\therefore f(36.1)=\sqrt {36.1}=f(36) +dy$
$\qquad=6+0.00833$
$\qquad=6.00833$

answered Aug 12, 2013 by