1.

Use differentials to find the approximate value of `(log)_e(4. 01),`having given that `(log)_e4=1. 3863`.

Answer» let `y= f(x) = log_e x`
`x + /_ x = 4.01`
`x + /_ x = 4 + 0.01`
`x= 4 & /_ x = 0.01`
`y= f(4) = log_e 4 = 1.3863`
`/_ x = 0.01`
`del x = 0.01`
`dx = 0.01`
now, `y= log_e x`
diff wrt x
`dy/dx = 1/x`
at x=4`dy/dx = 1/4`
`dy = dy/dx xx dx = 1/4 xx 0.01 = 0.0025`
`/_ y = 0.0025`
`log_e (4.01) = y + /_y `
`= 1.3868 + 0.0025`
`= 1.3888`
`log_e (4.01) = 1.3838`
Answer


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