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Solution for Using Differential, Find the Approximate Value of the √ 401 ? - CBSE (Science) Class 12 - Mathematics

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Question

Using differential, find the approximate value of the \[\sqrt{401}\] ?

Solution

\[\text { Consider the function y } = f\left( x \right) = \sqrt{x} . \]

\[\text { Let }: \]

\[ x = 400 \]

\[x + ∆ x = 401\]

\[\text { Then }, \]

\[ ∆ x = 1\]

\[\text { For } x = 400, \]

\[ y = \sqrt{400} = 20\]

\[\text { Let }: \]

\[ dx = ∆ x = 1\]

\[\text { Now,} y = \sqrt{x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 400} = \frac{1}{40}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{40} \times 1 = \frac{1}{40}\]

\[ \Rightarrow ∆ y = \frac{1}{40} = 0 . 025\]

\[ \therefore \sqrt{401} = y + ∆ y = 20 . 025\]

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Solution for question: Using Differential, Find the Approximate Value of the √ 401 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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