ANNE 發問於 科學及數學數學 · 10 年前

1 calculus Q.

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1 個解答

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  • 10 年前
    最愛解答

    I assume you require a differentiation of y with respect to x.

    y = x sqrt[(1 - x)/(1 + x)]

    By product rule,

    dydx = (dx/dx) sqrt[(1 - x)/(1 + x)] + x d{sqrt[(1 - x)/(1 + x)]} / dx

    = sqrt[(1 - x)/(1 + x)] + x d{sqrt[(1 - x)/(1 + x)]} / d[(1 - x)/(1 + x)]‧d[(1 - x)/(1 + x)]/dx (Chain rule)

    = sqrt[(1 - x)/(1 + x)] + x{1/2sqrt[(1 - x)/(1 + x)]}‧{-(1 + x) - (1 - x)}/(1 + x)^2 (Quotient rule)

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

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

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

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

    資料來源: Prof. Physics
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