How do you find the derivative of y= √X^2 - 1?
I am not exactly sure if I did this problem correctly. The answer I got is y= x/x^2-1^1/2. The problem I am having with this is the 2x. I am not sure if the 2 would cancel because of the 1/2 and just leave x by itself?
This is what I have done so far:
y = √x^2-1
y =(x^2 -1)^1/2
y=(1/2(x^2-1)^-1/2) 2x
y= x(x^2-1)^-1/2
y= x/x^2-1^1/2
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Verified answer
To find derivative of ---
=> y = √[ x² - 1 ]
=> y' = [ (1/2) ( x²-1 )^(- 1/2 ) ] * [ d/dx (x² - 1) ]
. . . . . . . 1 . . . .. . 1
=> y' = ------ * -------------- x ( 2 x - 0 )
. . . . . . . 2 . . √[ x² - 1 ]
. . . . . .. . 1 . . . . . . 1
=> y' = ------ x -------------- x 2 x
. . . . . . . . 2 . . √[ x² - 1 ]
Note in this case 2 in the numerator ( with x will get cancelled with the 2 in denominator ).
. . . . . . .. . .x
=> y' = --------------- ........................ Answer Answer
. . . . . . .√[ x² - 1 ]
>>>>>>>>>>>>>>>>>
Note : I have noticed that most of the students dont know the use of proper brackets. This habit creates a lot of confusion.
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