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Application Of Derivatives

Question

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Use differentials to approximate fourth root of 82.

Solution

Take space straight y space equals space straight x to the power of 1 fourth end exponent comma space space space straight x space equals space 81 comma space space space space dx space equals space δx space equals space 1 space space space so space that space straight x space plus space δx space equals space 82 Now space space space straight y space plus space δy space equals left parenthesis straight x plus δx right parenthesis to the power of 1 fourth end exponent space space space space space 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 fourth end exponent space minus space straight y space equals left parenthesis 82 right parenthesis to the power of 1 fourth end exponent minus 3

rightwards double arrow space space space space left parenthesis 82 right parenthesis to the power of 1 fourth end exponent space equals space δy space plus space 3

...(1)
Now

δy

is approximately equal to dy.

and space space dy space equals space dy over dx dx space equals space 1 fourth straight x to the power of negative 3 over 4 end exponent dx space equals space fraction numerator 1 over denominator 4 straight x to the power of begin display style 3 over 4 end style end exponent end fraction dx space equals space fraction numerator 1 over denominator 4 left parenthesis 81 right parenthesis to the power of begin display style 3 over 4 end style end exponent end fraction space left parenthesis 1 right parenthesis space equals space fraction numerator 1 over denominator 4 cross times 27 end fraction space equals space 1 over 108 space equals space 0.009 therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space space left parenthesis 82 right parenthesis to the power of 1 fourth end exponent space equals space 0.009 plus 3 space equals space 3.009.

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