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Application of Derivatives

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Short Answer Type

371. Use differentials to approximate:

open parentheses 25 close parentheses to the power of 1 third end exponent

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372. Use differentials to approximate:

left parenthesis 26.57 right parenthesis to the power of 1 third end exponent

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373. Use differentials to approximate:
cube root of 26

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374. Use differentials to approximate:
cube root of 63

Take $straight y space equals space straight x to the power of 1 third end exponent comma space space space space straight x space equals space 64 comma space space space space dx space equals space δx space equals space minus 1 space space space space so space that space space straight x space plus space δx space equals space 63$
$Now comma space space space straight y plus space δy space equals space left parenthesis straight x plus δx right parenthesis to the power of 1 third end exponent rightwards double arrow space space space space space δy space equals space left parenthesis straight x plus δx right parenthesis to the power of 1 third end exponent space minus space straight y space equals space left parenthesis 63 right parenthesis to the power of 1 third end exponent space minus space left parenthesis 64 right parenthesis to the power of 1 third end exponent space minus space 4$
$rightwards double arrow space space space space space space left parenthesis 63 right parenthesis to the power of 1 third end exponent space equals space δy space plus space 4$ ...(1)
Now, $δy$ is approximately equal to dy
$and space dy space equals space dy over dx dx space equals space 1 third straight x to the power of negative 2 over 3 end exponent space dx space equals space fraction numerator 1 over denominator 3 straight x to the power of begin display style 2 over 3 end style end exponent end fraction dx space equals space fraction numerator 1 over denominator 3 left parenthesis 64 right parenthesis to the power of begin display style 2 over 3 end style end exponent end fraction left parenthesis negative 1 right parenthesis space equals fraction numerator space minus 1 over denominator 3 cross times space 16 end fraction space equals space minus 1 over 48 therefore space space space from space left parenthesis 1 right parenthesis comma space space left parenthesis 63 right parenthesis to the power of 1 third end exponent space equals space minus 1 over 48 plus 4 space equals space fraction numerator negative 1 plus 192 over denominator 48 end fraction equals 191 over 48.$