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September 16, 2014

September 16, 2014

Posted by **Terry** on Sunday, June 5, 2011 at 4:09pm.

- calculus -
**Henry**, Sunday, June 5, 2011 at 4:56pmF(x) = Y = 4/sqrt(X) = 4/X^(1/2),

Y = 4X^(-1/2),

Y'=-2X^(-1/2-1) = -2X^(-3/2= -2/X^(3/2)= -2/sqrt(X^3).

- calculus -
**Terry**, Sunday, June 5, 2011 at 5:50pmusing the limit process how do i find the derivative of f(x)=4/sqrt of x?

- calculus -
**Reiny**, Sunday, June 5, 2011 at 6:08pmf '(x) = lim [4/√(x+h) - 4/√x]/h as h ---> 0

= lim [(4√x - 4√(x+h))/(√x√(x+h))*(1/h)*(4√x + 4√(x+h))/(4√x + 4√(x+h))

= lim [16x - 16(x+h)]/(√x√(x+h)(4√x + 4√(x+h)) * 1/h

= lim (-16h)/(√x√(x+h)(4√x + 4√(x+h)) * 1/h as h ---> 0

= -16/(2x(8√x))

= -1/(x√x) or -1/x^(3/2) or -(x^-(3/2)) or -1/(√x)^3

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