Use differentials to approximate $square root of 25.3 end root.$

Solution

Take $straight y equals space square root of straight x comma space space space straight x space equals space 25 space comma space space space dx space equals space δx equals space 0.3 space space so space that space straight x plus δx space equals space 25.3$
Then $δy space equals space square root of 25.3 end root space minus space square root of 25$ $open square brackets because space space space δy space equals space square root of straight x plus δx end root space minus square root of straight x close square brackets space$
$therefore space space space space square root of 25.3 end root space space equals space 5 space plus space δy$ ...(1)
Now, $δy$ is approximately equal to dy and
$dy space equals dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction left parenthesis 0.3 right parenthesis space equals space fraction numerator 1 over denominator 2 square root of 25 end fraction left parenthesis 0.3 right parenthesis space equals space fraction numerator 0.3 over denominator 10 end fraction space equals space 0.03$
$therefore space space space from space left parenthesis 1 right parenthesis comma space space space square root of 25.3 end root space equals space 5 plus 0.03 space equals space 5.03$

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Application Of Derivatives

Use differentials to approximate $square root of 25.3 end root.$

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