Question

approximate √36.03 by using the concept of derivative. anyone?

Answer 1

Let x = 36, so y =6 from \[\large y = \sqrt x\] so \[\Large \frac{ dy }{ dx } = \frac{ 1 }{ 2 \sqrt x }\]
\[\Large dy = \frac{ 1 }{ 2 \sqrt x } dx\]
\[\Large \Delta y = \frac{ 1 }{ 2 \sqrt x } \Delta x\]
Again, x =36 so let the Δx be 0.03. Find Δy, then you can approximate the sqrt of 36.03 by adding Δy to y (which is 6).

Answer 2

http://www.mathwords.com/a/approximation_by_differentials.htm

Answer 3

how to find dx? @agent0smith

Answer 4

If x is 36, dx is 0.03, since we want to approximate sqrt 36.03, so 36.03-36 = 0.03.

Answer 5

dx just means the change in x. x changes from 36 to 36.03, and we use 36 as x since we know the sqrt of 36 is 6.

Answer 6

so it was like this? @agent0smith