a trick is involved. Realize that
|x|=sqrt(x^2)
y=sqrt(x^2+3x+4)
y'=1/(2*sqrt( )) * (2x+3)/1
Y=|X^2+3x-4|
find dy/dx?
3 answers
Following bob's own suggestion, it should have been:
y = √ ( (x^2 + 3x + 4)^2)
= [( (x^2 + 3x + 4)^2)]^(1/2)
dy/dx = (1/2) ( (x^2 + 3x + 4)^2)^(-1/2) (2)(x^2 + 3x + 4) (2x+3)
= (2x+3)√(x^2 + 3x + 4)
or
we could do it piece-meal
y=|x^2+3x-4|
--> y = x^2 + 3x - 4 , for x <-4 OR x > 1
dy/dx = 2x + 3 for x <-4 OR x > 1
or
y = -x^2 - 3x + 4 for -4 < x < 1
dy/dx = -2x - 3 for -4 < x < 1
dy/dx is not defined at x = -4 or x = 1
y = √ ( (x^2 + 3x + 4)^2)
= [( (x^2 + 3x + 4)^2)]^(1/2)
dy/dx = (1/2) ( (x^2 + 3x + 4)^2)^(-1/2) (2)(x^2 + 3x + 4) (2x+3)
= (2x+3)√(x^2 + 3x + 4)
or
we could do it piece-meal
y=|x^2+3x-4|
--> y = x^2 + 3x - 4 , for x <-4 OR x > 1
dy/dx = 2x + 3 for x <-4 OR x > 1
or
y = -x^2 - 3x + 4 for -4 < x < 1
dy/dx = -2x - 3 for -4 < x < 1
dy/dx is not defined at x = -4 or x = 1
wolframalpha cleverly combines the two solutions as
(2x-3) * y/|y|
which takes care of the +/- cases.
(2x-3) * y/|y|
which takes care of the +/- cases.
1 answer