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Application of Derivatives

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Short Answer Type

361.

Use differentials to approximate:
$square root of 3.968 end root$

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362.

Use differentials to approximate:
$square root of 25.2 end root$

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363.
Use differentials to approximate:
$left parenthesis 0.0037 right parenthesis to the power of 1 half end exponent$

Take $straight y space equals space square root of straight x comma space space space space straight x space space equals 0.0036 space space space dx space equals space δx space equals space 0.0001 space space so space that space space straight x space plus space δx space equals space 0.0037$
$Now space space space space space space space space space straight y space plus space δy space equals space square root of straight x plus δx end root$
$rightwards double arrow space space space space space δy space equals space square root of straight x plus δx end root space minus space straight y space equals space square root of.0037 end root space minus space square root of.0036 end root space space space space space rightwards double arrow space space space space space space δy space equals space square root of.0037 end root space minus space.06 rightwards double arrow space space space space space square root of.0037 end root space equals space δy space plus space.06 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space space space space space space δy space is space approximately space equal space to space dy space space space space space space space$
$and space dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx space equals space fraction numerator 1 over denominator 2 square root of 0.0036 end root end fraction left parenthesis 0.0001 right parenthesis space equals fraction numerator 0.0001 over denominator 2 left parenthesis 0.06 right parenthesis end fraction space equals space fraction numerator 0.0001 over denominator 0.12 end fraction space equals space fraction numerator 0.01 over denominator 12 end fraction space space space space space space equals space 0.000833 space space space therefore space space space from space left parenthesis 1 right parenthesis comma space space space square root of 0.0037 end root space equals space 0.000833 space plus space 0.6 space equals space 0.060833$