+91-8076753736 [email protected]

-->

Application Of Derivatives

Question

Wired Faculty App

Use differentials to approximate:
$left parenthesis 26.57 right parenthesis to the power of 1 third end exponent$

Solution

Let space space space straight y space equals space straight x to the power of 1 third end exponent comma space space space space straight x space equals space 27 comma space space dx space equals space minus 0.43 space space space space

δy space equals space left parenthesis straight x plus dx right parenthesis to the power of 1 third end exponent space space minus space straight x to the power of 1 third end exponent space equals space left parenthesis 26.57 right parenthesis to the power of 1 third end exponent space minus space left parenthesis 27 right parenthesis to the power of 1 third end exponent space equals space left parenthesis 26.57 right parenthesis to the power of 1 third end exponent space minus space 3

therefore space space space space space left parenthesis 26.57 right parenthesis to the power of 1 third end exponent space equals space 3 space plus space δy

...(1)
Now

δy

is approximately equal to dy

and space space dy space equals space dy over dx dx space equals space 1 third straight x to the power of negative 2 over 3 end exponent cross times dx space equals space fraction numerator 1 over denominator 3 straight x to the power of negative begin display style 2 over 3 end style end exponent end fraction cross times dx space equals space fraction numerator 1 over denominator 3 left parenthesis 27 right parenthesis to the power of negative begin display style 2 over 3 end style end exponent end fraction cross times space left parenthesis negative 0.43 right parenthesis

equals space fraction numerator 1 over denominator 3 cross times 9 end fraction cross times left parenthesis negative 0.43 right parenthesis space equals space minus fraction numerator 0.43 over denominator 27 end fraction space equals space minus 0.0159

therefore space space space space space space δy space equals space minus 0.0159 therefore space space from space left parenthesis 1 right parenthesis comma space space space left parenthesis 26.57 right parenthesis to the power of 1 third end exponent space equals space 3 space minus space 0.0159 space equals space 2.9841 space equals space 2.984

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation