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Application Of Derivatives

Question
CBSEENMA12035593

Use differentials to approximate:
square root of 26












Solution

 Take space straight y space equals space square root of straight x comma space space space straight x space equals space 25 comma space space space space dx space equals space δx space equals space 1 space space so space that space straight x space space plus δx space equals space 26
Now comma 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 square root of straight x space equals space square root of 26 space space minus 5
rightwards double arrow space space space space space space square root of 26 space equals space δy space plus space 5 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 δy space is space approximately space equal space to space dy
  and space space space dy space equals space dy over 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 cross times 5 end fraction cross times 1 space equals space 1 over 10 space equals space 0.1
therefore space space space from space left parenthesis 1 right parenthesis comma space space square root of 26 space equals space 0.1 plus 5 space equals space 5.01.