Giải Khám phá 3 trang 44 Toán 11 tập 2 Chân trời

$y'(x_{0}) =\lim_{x \to x_{0}}\frac{sinx-sinx_{0}}{x-x_{0}}$

Gọi $x = x_{0} + \Delta x$

Suy ra:

$y'(x_{0}) =\lim_{\Delta x \to 0}\frac{sin(x_{0}+\Delta x) -sinx_{0}}{\Delta x}$

$= \lim_{\Delta x \to 0}\frac{sinx_{0}cos\Delta x + cosx_{0}sin\Delta x -sinx_{0}}{\Delta x}$

$= \lim_{\Delta x \to 0}\frac{sinx_{0}cos\Delta x -sinx_{0}}{\Delta x} + \lim_{\Delta x \to 0}\frac{cosx_{0}sin\Delta x}{\Delta x}$

$= sinx_{0}\lim_{\Delta x \to 0}\frac{cos\Delta x - 1}{\Delta x}+cosx_{0}.\lim_{\Delta x \to 0}\frac{sin\Delta x}{\Delta x}$

Ta có: $\lim_{x \to 0}\frac{sinx}{x}=1$. Suy ra:$\lim_{\Delta x \to 0}\frac{sin\Delta x}{\Delta x}=1$

Ta lại có: $\lim_{\Delta x \to 0}\frac{cos\Delta x - 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{(cos\Delta x - 1)(cos\Delta x +1)}{\Delta x.(cos\Delta x +1)}$

$= \lim_{\Delta x \to 0}\frac{cos^{2}\Delta x - 1}{\Delta x.(cos\Delta x +1)} = -\lim_{\Delta x \to 0}\frac{sin^{2}\Delta x}{\Delta x.(cos\Delta x +1)}$

$= -\lim_{\Delta x \to 0}\frac{sin\Delta x}{\Delta x}.\lim_{\Delta x \to 0}\frac{sin\Delta x}{cos\Delta x+1} = 1.\frac{0}{1+1}=0$

Từ đó: $y'(x_{0}) = cosx_{0}.1=cosx_{0}$