f(x) = (sqrt((1-x)/(1+x)))

Question

f'(x)=?

f'(x)=o

x=??

f''(x)=??

f''(x)=o

x=??

Answer 1

f(x) = Sqrt[(1-x)/(1+x)]

= [(1-x)/(1+x)]^1/2

= (1-x)^1/2/(1+x)^1/2

= (1-x)^1/2(1+x)^-1/2

Apply derivative on each side.

Remember the formula d/dx(uv) = uv'+vu'

f'(x) = (1-x)^1/2d/dx(1+x)^-1/2+(1+x)^-1/2d/dx(1-x)^1/2

= (1-x)^1/2(-1/2(1+x)^-3/2)+(1+x)^-1/2(1/2(1-x)^-1/2

=-1/2 (1-x)^1/2(1+x)^-3/2+1/2(1+x)^-1/2(1-x)^-1/2

f'(x) = [-(1-x)^1/2(1+x)^-3/2]/2+[(1+x)^-1/2(1-x)^-1/2]/2

At x = 1

f'(x) = 0

To find f''(x) again derivative to each side.

f''(x) = d/dx[-1/2 (1-x)^1/2(1+x)^-3/2+1/2(1+x)^-1/2(1-x)^-1/2

= d/dx[-1/2 (1-x)^1/2(1+x)^-3/2]+d/dx[1/2(1+x)^-1/2(1-x)^-1/2]

= -1/2[(1-x)^1/2(-3/2)(1+x)^-5/2+(1+x)^-3/2(1/2)(1-x)^-1/2]+1/2[(1+x)^-1/2(-1/2)(1-x)^-3/2+        (1-x)^-1/2(-1/2)(1+x)^-3/2]

= 1/2(1-x)^1/2(1+x)^-5/2-1/4(1+x)^-3/2(1-x)^-1/2-1/4(1+x)^-1/2(1-x)^-3/2

   -1/4(1-x)^-1/2(1+x)^-3/2

= 1/2(1-x)^(1/2)(1+x)^(-5/2)-1/2(1+x)^(-3/2)(1-x)^(-1/2)-1/4(1+x)^(-1/2)(1-x)^(-3/2)

At x = 1 f''(x) = 0