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  • Using differentials, find the approximate value of  up to two places of decimals.    <

Using differentials, find the approximate value of \sqrt{ 25.3} up to two places of decimals.

 

 
 
 
 
 

Answers (1)

Assume 

f(x) =\sqrt{x}

f'(x) =\frac{1}{2\sqrt{x}}

f(x+\Delta x) =f(x)+\Delta x f'(x)

f(25.3) =f(25)+ 0.3 f'(25)

\sqrt{25.3}= f(25.3) =5+ 0.3 \times \frac{1}{2 \times 5} = 5.03

Posted by

Safeer PP

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