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Application Of Derivatives

Question

Wired Faculty App

Use differentials to approximate fourth root of 15.

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 16 comma space space space dx space equals space δx space equals space minus 1 space space space so space that space straight x plus space δx space equals space 15 Now space space space straight y space plus space δy space space equals space left parenthesis straight x plus δx right parenthesis to the power of 1 fourth end exponent space space space space space space space space rightwards double arrow space 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 space left parenthesis 15 right parenthesis to the power of 1 fourth end exponent minus 2 rightwards double arrow space space space space left parenthesis 15 right parenthesis to the power of 1 fourth end exponent space equals space δy space plus 2 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 δy space is space approximately space equal space to space dy

and 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 16 right parenthesis to the power of begin display style 3 over 4 end style end exponent end fraction left parenthesis negative 1 right parenthesis space equals space minus 1 over 32 equals negative 0.0312 therefore space space space space from space left parenthesis 1 right parenthesis comma space space space left parenthesis 15 right parenthesis to the power of 1 fourth end exponent space equals space minus 0.0312 plus 2 space equals space 1.9688.

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