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calculus

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how do I find the derivative of f(x)= 4/ square root of x ? using the limit process?

  • calculus - ,

    F(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 - ,

    using the limit process how do i find the derivative of f(x)=4/sqrt of x?

  • calculus - ,

    f '(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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