Question

两道关于导数的题目

1，求y=3x^3/sinx-cosx的导数
2，设y=1-√x / 1+√x ,求y'(4)

要详细的求解过程，谢谢了！

Answer 1

(1)
y = 3x³/(sinx - cosx)
y' = 3 • [(sinx - cosx)(3x²) - (x³)(cosx + sinx)]/(sinx - cosx)²
= 3(3x²sinx - 3x²cosx - x³cosx - x³sinx)/(sinx - cosx)²
= 3x²(3sinx - 3cosx - xcosx - xsinx)/(sinx - cosx)²
(2)
y = (1 - √x)/(1 + √x)
y' = [(1 + √x)(-1/2√x) - (1 - √x)(1/2√x)]/(1 + √x)²
= [(1 + √x)(-1) - (1 - √x)]/[2√x(1 + √x)²]
= (- 1 - √x - 1 + √x)/[2√x(1 + √x)²]
= - 1/[√x(1 + √x)²]
y'|(4) = - 1/[√4(1 + √4)²] = -1/[(2)(3)²] = -1/18

Answer 2

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