Question

Find the approximate values, using differentials :
• √37
• √401

Answer should be appropriate ..!!​

Answer 1

Answer:

put so means .•. in every point like 1 ) pair .

Answer 2

Question :

• Find the approximate values, using differentials : • √37 • √401

Solution :

$\longmapsto\bf Solving\:for\:\pink{\sqrt{37}}$

Let y = f(x) = √x

Take x = 36 ,

$\leadsto \sf x + \Delta x = 37 \\\dashrightarrow \Delta x = 37 - x \\\dashrightarrow \Delta x = 1$

When x = 36 , then y = √36 = 6

Let dx = ∆x = 1

Now,

y = √x so that $\sf\dfrac{dy}{dx} =\dfrac{1}{2\sqrt{x}}$

therefore, $\sf\green{\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{36}} = \dfrac{1}{12}} = 0.83$

$\longmapsto\sf dy=\dfrac{dy}{dx}.dx$

$\dashrightarrow \sf dy = \dfrac{1}{12}(1) = 0.83$

$\sf\implies\blue{ \Delta y = 0.83 }$ $\quad\because\{\Delta y \sf{\:is\: approx.}= dy \}$

$\longmapsto\sf \pink{\sqrt{37}}$

$\longmapsto \sf y + \Delta y = 6 + 0.83$

$\dashrightarrow 6.083$

Hence,

• √37 = 6.083

╼╼╼╼╼╼╼╼╼╼╼╼

╼╼╼╼╼╼╼╼╼╼╼╼

$\longmapsto\bf Solving\:for\:\pink{\sqrt{401}}$

Let y = f(x) = √x

Take x = 400 ,

$\leadsto \sf x + \Delta x = 401 \\\dashrightarrow \sf\Delta x = 401 - 400 \\\dashrightarrow \sf\Delta x = 1$

When x = 400 , then y = √400 = 20

Let dx = ∆x = 1

Now,

y = √x so that $\sf\dfrac{dy}{dx} =\dfrac{1}{2\sqrt{x}}$

therefore, $\sf\green{\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{400}} = \dfrac{1}{40}} = 0.025$

$\longmapsto\sf dy=\dfrac{dy}{dx}.dx$

$\dashrightarrow \sf dy = 0.025(1) = 0.025$

$\implies\sf\blue{ \Delta y = 0.025 }$

$\quad\because \{ \Delta y \sf{\:is\: approx.}= dy \}$

$\longmapsto\sf \pink{\sqrt{401}}$

$\longmapsto \sf y + \Delta y = 20 + 0.025$

$\dashrightarrow 20.025$

Hence,

• √401 = 20.025