of the square root of (x^2 - 4)? could you show steps please?
i swear calculus is going to drive me insaneeee
2007-09-22 17:53:01 · 5 answers · asked by Anonymous 3 in Science & Mathematics ➔ Mathematics
y = (x ² - 4) ^(1/2)
dy / dx = (1/2) (x ² - 4) ^(- 1/2) (2x)
dy / dx = x / (x ² - 4) ^(1/2)
OR
y = u ^(1/2)---------- u = (x ² - 4)-----du/dx = 2x
dy / du
= (1/2) u ^(-1/2)
= 1 / [ 2 u^(1/2) ]
= 1 / [ 2 (x ² - 4) ^(1/2) ]
dy / dx = [ dy /du ] [ du / dx ]
dy / dx = ( x ) / [ (x ² - 4)^(1/2) ]
2007-09-22 22:07:24 · answer #1 · answered by Como 7 · 0⤊ 1⤋
>> d/dx (x^2 – 4)^1/2
use x^n = nx^n-1
2x
________
2âx^2 – 4
= x (numerator)
________
âx^2 - 4 (denominator)
2007-09-23 01:59:13 · answer #2 · answered by meeta1704 2 · 0⤊ 1⤋
(x^2 - 4)^1/2
1/2(x^2-4)^-1/2 * 2x
= x
------
sqrt(x^2-4)
2007-09-23 01:01:05 · answer #3 · answered by Axis Flip 3 · 0⤊ 0⤋
y = (x^2 - 4)^1/2
dy/dx = (1/2)(x^2 - 4)^[-1/2](2x)
dy/dx = x/(x^2 - 4)^1/2
For example, d(u^1/2)/dx = (1/2)u^(-1/2)(du)
2007-09-23 01:00:01 · answer #4 · answered by Anonymous · 0⤊ 1⤋
y=sqrt(x^2-4)
y'=1/2(x^2-4)^(-1/2) *2x
y'=x/(x^2-4)^1/2
2007-09-23 01:02:32 · answer #5 · answered by ptolemy862000 4 · 0⤊ 1⤋