返回题库

导数：

Derive from first principles

专题
General / 综合
难度
L4

题目详情

从第一原理出发，推导以下函数的导数

g(x)=ecos⁡(x)g(x) = e^{\cos (x)}

英文原题

Derive, from first principles, the derivative of

g(x)=ecos⁡(x)g(x) = e^{\cos (x)}
解析

由定义

g′(x)=lim⁡h→0ecos⁡(x+h)−ecos⁡xh.g'(x)=\lim_{h\to 0}\frac{e^{\cos(x+h)}-e^{\cos x}}{h}.

将差分因式分解：

ecos⁡(x+h)−ecos⁡xh=ecos⁡x⋅ecos⁡(x+h)−cos⁡x−1h.\frac{e^{\cos(x+h)}-e^{\cos x}}{h}=e^{\cos x}\cdot\frac{e^{\cos(x+h)-\cos x}-1}{h}.

又 cos⁡(x+h)−cos⁡x=−sin⁡x h+o(h)\cos(x+h)-\cos x=-\sin x\,h+o(h)，并且 lim⁡u→0eu−1u=1\lim_{u\to 0}\frac{e^u-1}{u}=1，因此

g′(x)=−sin⁡x ecos⁡x.\boxed{g'(x)=-\sin x\,e^{\cos x}}.

英文解析

By definition,

g′(x)=lim⁡h→0ecos⁡(x+h)−ecos⁡xh.g'(x)=\lim_{h\to 0}\frac{e^{\cos(x+h)}-e^{\cos x}}{h}.

Factor the difference:

ecos⁡(x+h)−ecos⁡xh=ecos⁡x⋅ecos⁡(x+h)−cos⁡x−1h.\frac{e^{\cos(x+h)}-e^{\cos x}}{h}=e^{\cos x}\cdot\frac{e^{\cos(x+h)-\cos x}-1}{h}.

Since cos⁡(x+h)−cos⁡x=−sin⁡x h+o(h)\cos(x+h)-\cos x=-\sin x\,h+o(h) and lim⁡u→0eu−1u=1\lim_{u\to 0}\frac{e^u-1}{u}=1, it follows that

g′(x)=−sin⁡x ecos⁡x.\boxed{g'(x)=-\sin x\,e^{\cos x}}.