calc help....please!

[withy]

i'm working on some on my calc homework and i'm having a little trouble with this problem in particular:
the demand function is modeled by:

P = 740-x^1/2 , 0 <=x <= 740

find the interval on which the demand is elastic, inelastic, and of unit elasticity.

price elasticity = (p/x)/(dp/dx)

so i know that i have to substitute (740-x)^1/2 into the equation..... but how?

i e-mailed the math lab hear and they gave me this:

(((740-x)^1/2)/x)/(-1/(2/((740-x)^1/2)))

so my question to everyone is why is d/dx = (-1/(2/((740-x)^1/2)))?!?!?!?!?

its a simple problem but i jsut can't see it....

[projekt_s13]

ummmmm........... yea......... Sorry i cant help you there. o_O

[withy]

ehhh, thanks for looking.

[ManoNegra]

P is a function of x so -> p(x) = 740 - x^1/2

derivative of p(x) = p'(x) = -(1/2)x^(-1/2) = -1/[2x^1/2]

p/x = (740 - x^1/2) / x

so (p/x)/(dp/dx) = [(740 - x^1/2) / x ] / [-1/(2x^1/2)]

simplify to get this:

price elasticity = 1/2 - 370/x^1/2 = e(x) for simplicity

e'(x) = 160 / x^3/2 derivative of price elasticity function

You can graph this function to find out the interval of growth (elasticity?), decline (inelastic?) and unity (equate the function to 1 and solve for x).

Been a while so make sure you double check your work with a classmate or teacher.