Question

if y= logx where x=3 delta x=0.03 then the absolute error in y is​

Answer 1

Answer:

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Answer 2

The absolute error in y is​ 0.01.

Explanation:

Given : y= log x  , where x=3  and $\Delta x=0.03$

Differentiate both sides of the above function with respect to x , we get

$\dfrac{dy}{dx}=\dfrac{1}{x}$

$\Rightarrow\ \dfrac{\Delta y}{\Delta x}=\dfrac{1}{x}\\\\\Rightarrow\ \Delta y= \dfrac{1}{x}\times\Delta x$

Substitute the value of x= 3 and $\Delta x=0.03$ , we get

$\Delta y= \dfrac{1}{3}\times(0.03)\\\\\Rightarrow\ \Delta y=0.01$

Hence, the absolute error in y is​ 0.01 .

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