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CBSE Class 12-science Answered

Find the value of left parenthesis 244 right parenthesis to the power of bevelled 2 over 5 end exponent approximately using differentials.
Asked by Topperlearning User | 05 Aug, 2014, 12:38: PM
Expert Answer

 Take space straight x space as space 243 space and space increment straight x equals 1
Let space straight y equals straight x to the power of bevelled 2 over 5 end exponent
Then comma
increment straight y equals left parenthesis straight x plus increment straight x right parenthesis to the power of bevelled 2 over 5 end exponent minus straight x to the power of bevelled 2 over 5 end exponent
rightwards double arrow increment straight y equals left parenthesis 244 right parenthesis to the power of bevelled 2 over 5 end exponent minus 9
rightwards double arrow left parenthesis 244 right parenthesis to the power of bevelled 2 over 5 end exponent equals increment straight y plus 9
Now comma
dy equals open parentheses dy over dx close parentheses increment straight x
space space space space equals open parentheses 2 over 5 straight x to the power of minus bevelled 3 over 5 end exponent close parentheses dx
space space space space equals 2 over 5 open parentheses 243 close parentheses to the power of minus bevelled 3 over 5 end exponent cross times 1
space space space space equals fraction numerator 2 over denominator 5 cross times 27 end fraction
space space space space equals 0.014
Therefore comma
open parentheses 244 close parentheses to the power of bevelled 2 over 5 end exponent equals 9 plus 0.014

Answered by | 05 Aug, 2014, 02:38: PM
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