Question

一个简单导数公式应该怎么样推导，如f(x)/g(x)

Answer 1

楼上都答非所问，问的是「推导过程」，可从定义出发。
导数的商公式证明：
设有函数u = f(x)及v = g(x)

(u/v)' = d/dx [f(x)/g(x)] = lim(Δx→0) Δy/Δx

= lim(Δx→0) [f(x + Δx)/g(x + Δx) - f(x)/g(x)]/Δx

= lim(Δx→0) 1/[g(x + h)g(x)] • lim(Δx→0) [f(x + Δx)g(x) - f(x)g(x + Δx)]/Δx

= 1/[g(x + 0)g(x)] • lim(Δx→0) {[f(x + Δx)g(x) - f(x)g(x)] - [f(x)g(x + Δx) - f(x)g(x)]}/Δx，加减f(x)g(x)

= 1/[g(x)]² • lim(Δx→0) {[f(x + Δx)g(x) - f(x)g(x)]/Δx - [f(x)g(x + Δx) - f(x)g(x)]/Δx}

= 1/[g(x)]² • {g(x) • lim(Δx→0) [f(x + Δx) - f(x)]/Δx - f(x) • lim(Δx→0) [g(x + Δx) - g(x)]/Δx}

= 1/[g(x)]² • [g(x) • d/dx f(x) - f(x) • d/dx g(x)]

= [g(x)f'(x) - f(x)g'(x)]/[g(x)]²

= (vu' - uv')/v²

Answer 2

f(x)±g(x)的导数是 f'(x)±g'(x)
f(x)/g(x) [f'(x)g(x)-g‘(x)f(x)]/[g(x)]²
f(x)g(x)=f'(x)g(x)+f(x)g'(x)

Answer 3

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