Use differentials to approximate:
$square root of 25.2 end root$

Solution

Take $straight y equals space square root of straight x comma space space space space straight x space equals space 25 comma space space space space dx space equals space 0.2 space space so space that space straight x plus dx space equals space 25.2$
Then $δy space equals space square root of 25.2 end root space minus space square root of 25 space equals space square root of 25.2 space end root space minus space 5 space space space space space space space space space space space space space space space space space space space space space space open square brackets space because space space space δy space equals space square root of straight x plus δx end root space minus space square root of straight x close square brackets$
$therefore space space space space space square root of 25.2 end root space equals space 5 plus δy$ ...(1)
Now, $δy$ is approximately equal to dy and
$dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction left parenthesis 0.2 right parenthesis space equals space fraction numerator 1 over denominator 2 square root of 25 end fraction left parenthesis 0.2 right parenthesis space equals space fraction numerator 0.2 over denominator 10 end fraction space equals space 0.02$
$therefore space space space space from space left parenthesis 1 right parenthesis comma space space space square root of 25.2 end root$ is approximately equal to 5.02.

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Application Of Derivatives

Use differentials to approximate:
$square root of 25.2 end root$

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.

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