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

- 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$

$\therefore f(36.1)=\sqrt {36.1}=f(36) +dy$

$\qquad=6+0.00833$

$\qquad=6.00833$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...