Question

use diffrential to approximate √25.3
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Answer 1

$\large \underline{\underline{ \red{\mathfrak{Question}}}} : - \\ \rightarrow \sf{Use\: \green{ differential} \: \red{ to \: approximate} \: \sqrt{25.3} } \\ \\ \large \underline{ \underline{ \purple{\mathfrak{Answer}}}} : - \\ \\ \rightarrow \: \red{ \boxed{\sqrt{25.3} = 5.03}} \: \\ \\ \large \mathfrak{\underline{\underline{ \green{Step - }by - \red{step} \: \blue{explanation}}}} : - \\ \\ \green{ let \: \: } \sf{ y \: = \sqrt{x} } \\ \sf{ \blue{ again \: let \:} \: x = 25 \: \: \purple{and \: \triangle x = 0.3}} \\ \\ \green{ \tt we \: have \: }\\ \implies \sf{ y = \sqrt{x} } \\ \\ \sf{ \red{ differentiate }\: with \: \green{ respect }\: to \: x} \\ \\ \implies \: \sf{ \pink{ \frac{dy}{dx} } = \green{\frac{1}{2 \sqrt{x} } }} \: \: \: \: \because \sf{ \: \: x = 25} \\ \\ \implies \boxed{\sf{\: \red{\frac{dy}{dx} } = \blue{\frac{1}{10} }}} \\ \\ yet \\ \sf{ \implies \triangle y = \green{ \frac{dy}{dx}} \red{ \triangle x} }\\ \\ \implies \sf{ \triangle y = \pink{ \frac{1}{10}} \green{ \times 0.3}} \\ \\ \implies \: \boxed{ \blue{ \sf{\triangle y} = 0.03 }} \\ \bf{and \: also} \\ \\ \sf{ \triangle y = \red{ f(x + \triangle x)} - f(x)} \\ \\ \implies \sf{0.03 = \green{ \sqrt{x + \triangle x} } - \sqrt{x} } \\ \\ \implies \: \green{ 0.03} = \pink{\sqrt{25 + 0.3} } - 5 \\ \\ \implies \: \sqrt{25.3} = \purple{ 0.03 + 5 }\\ \\ \implies \: \boxed{ \red{\sqrt{25.3} = 5.03}} \\ \\ \underline{\sf{ used \: formula}} : - \\ \\ \star \: \sf{ \triangle y = \red{ f(x + \triangle x)} - f(x)} \: \\ \: \\ \star \: \sf{ \triangle y = \green{ \frac{dy}{dx}} \red{ \triangle x} } \:$

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