derivative?

Question

what is the derivative of
y = [(x-2)/k]^(1/2)

k is a constant

Answer 1

Remember, use the product rule. Also, since we have 1/k inside the parentheses, we are also taking its square root. When we do that, we introduce a new constant: (1/k)^½.

First function (1/k)^½. times the derivative of the second, [½ (x - 2)^-½] x ', plus the product of the second function [(x - 2)]^½ times the derivative of the first, (0), because (1/k)^½. is a constant.

y = [(x-2)/k]^(1/2)
y = (1/k)^½ [(x - 2)]^½
y ' = (1/k)^½ [½ (x - 2)^-½] x ' + [(x - 2)]^½ (0)
y ' = (1/k)^½ [½ (x - 2)^-½] x ' + 0
y ' = (1/k)^½ [½ (x - 2)^-½] x '

Answer 2

This is a chain rule question, so deal with power first and then the inside of the brackets.

The other two answers have missed the 1/k out from differentiating the bracket

d/dx([(x-2)/k]^(1/2)) = (1/2).[(x-2)/k]^(-1/2)].(1/k)

Answer 3

y = [ ( x - 2 ) / k ] ^ ( 1 / 2 )

multiplying the whole equation by the power
and subtracting one from the power.
and multiplying the whole equation by the derivative of the abstract [ ( x - 2 ) / k ] , which is just 1.

y' = ( 1 / 2 ) . [ ( x - 2 ) / k ] ^ ( - 1 / 2 ) . ( 1 )
y' = 1 / 2 [ ( x - 2 ) / k ] ^ ( - 1 / 2 )

Answer 4

y'=[(1/k)^(1/2)]/2(x-2)^(1/2)