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Application Of Derivatives

Question
CBSEENMA12035608

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent

Solution
straight y space equals space straight x to the power of 1 fifth end exponent comma space space space space space straight x space equals space 32 comma space space space space dx space equals space 0.15
space δy space equals space left parenthesis straight x plus dx right parenthesis to the power of 1 fifth end exponent space space minus straight x to the power of 1 fifth end exponent space equals space left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent space minus space left parenthesis 32 right parenthesis to the power of 1 fifth end exponent space equals space left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent space minus space 2
therefore space space space space left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent space equals space 2 plus space δy 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 δy space is space approximate space value space of space dy
and space space space space space space space space dy space equals space dy over dx dy space equals space 1 fifth straight x to the power of negative 4 over 5 end exponent dx
space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1 over denominator 5 straight x to the power of begin display style 4 over 5 end style end exponent end fraction dx space equals space fraction numerator 1 over denominator 5 left parenthesis 32 right parenthesis to the power of begin display style 4 over 5 end style end exponent end fraction space cross times space left parenthesis 0.15 right parenthesis
space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 0.15 over denominator 5 space cross times space 16 end fraction space equals space fraction numerator 0.15 over denominator 80 end fraction space equals space 0.0018
therefore space space space from space left parenthesis 1 right parenthesis comma space space space left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent space equals space 2 plus space δy space equals space 2 plus 0.0018 space equals space 2.0018 space equals space 2.002