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Question

Using differentials, find the sum of  digits  approximate value of the following up to $$3$$ places of decimal.
$$(15)^{\tfrac{1}{4}}$$


Solution

 Consider $$y=x ^{\tfrac{1}{4}}$$.  

Let $$x = 16$$ and $$\Delta x = -1$$. 

Then,

$$\Delta y={(x+\Delta x)}^{\tfrac{1}{4}}-x^{\tfrac{1}{4}}=(15)^{\tfrac{1}{4}}-(16)^{\tfrac{1}{4}}=(15)^{\tfrac{1}{4}}-2$$ 

$$\Rightarrow (15)^{\tfrac{1}{4}}=2+\Delta y$$

Now, $$dy$$ is approximately equal to $$\Delta $$y and is given by,

$$dy=\left ( \dfrac{dy}{dx} \right )\Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)$$        [as $$y= x^{{\frac{1}{4}}}$$]

$$=\frac{1}{4(16)^{\frac{3}{4}}}(-1)=\dfrac{-1}{4 \times 8}=\dfrac{-1}{32}=-0.03125$$

Hence, the approximate value of $$(15)^{\frac{1}{4}}$$ 
is $$2 + (-0.03125) = 1.96875\approx1.968$$. 

Mathematics
RS Agarwal
Standard XII

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