Use differentials to approximate:
$square root of 0.6 end root$

Solution

Take $straight y space equals square root of straight x comma space space space space straight x space space equals 0.64 comma space space space dx space equals space δx space equals space minus 0.04 space so space that space straight x space plus space straight delta space straight x space equals space 0.6$
Now, $straight y plus δy space equals space square root of straight x plus δx end root space space space space space rightwards double arrow space space space δy space equals space square root of straight x plus δx end root space minus space straight y space equals space square root of 0.6 end root space space minus space 0.8$
$rightwards double arrow space space space space space space space space space space space space space square root of 0.6 end root space equals space δy space plus space 0.8$ ...(1)
Now, $δy$ is approximately equal to dy
and $dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx space equals space fraction numerator 1 over denominator 2 square root of 0.64 end root end fraction left parenthesis negative 0.04 right parenthesis$
$equals negative fraction numerator 0.04 over denominator 2 space cross times space 0.8 end fraction space equals space minus fraction numerator 0.04 over denominator 1.6 end fraction space equals space minus 0.025$
$therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space square root of 0.6 end root space equals space minus 0.025 plus 0.8 space equals space 0.775$

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Application Of Derivatives

Use differentials to approximate:
$square root of 0.6 end root$

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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