SOLUTION: Given that y = lm(x^2 - 4) for x > 2, find dy/dx. Hence, find, in terms of p , the approximate change in y when x increases from 3 to 3 + p , where p is small.
Algebra ->
Test
-> SOLUTION: Given that y = lm(x^2 - 4) for x > 2, find dy/dx. Hence, find, in terms of p , the approximate change in y when x increases from 3 to 3 + p , where p is small.
Log On
Question 1194171: Given that y = lm(x^2 - 4) for x > 2, find dy/dx. Hence, find, in terms of p , the approximate change in y when x increases from 3 to 3 + p , where p is small. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! y=ln (x^2-4)
y'=2x/(x^2-4)
when x=3, y=6/5 or 1.2 or 1.2 p change in y for every p change in x
