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प्रश्न
1 Using differential, find the approximate value of the following:
\[\sqrt{25 . 02}\]
बेरीज
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उत्तर
\[\text { Consider the function y } = f\left( x \right) = \sqrt{x} . \]
\[\text { Let }: \]
\[ x = 25 \]
\[ x + ∆ x = 25 . 02\]
\[\text { Then, } \]
\[ ∆ x = 0 . 02\]
\[\text { For} x = 25, \]
\[ y = \sqrt{25} = 5\]
\[\text { Let }: \]
\[ dx = ∆ x = 0 . 02\]
\[\text { Now,} y = \sqrt{x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]
\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 25} = \frac{1}{10}\]
\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002\]
\[ \Rightarrow ∆ y = 0 . 002\]
\[ \therefore \sqrt{25 . 02} = y + ∆ y = 5 . 002\]
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