Giải Khám phá 4 trang 44 Toán 11 tập 2 Chân trời

a) $y'(x_{0}) = \lim_{x \to x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}=\lim_{x \to x_{0}}\frac{e^{x}-e^{x_{0}}}{x-x_{0}}$

Gọi $x = x_{0} + \Delta x$

Suy ra: $y(x_{0})' = \lim_{\Delta x \to 0}\frac{e^{x_{0}+\Delta x}-e^{x_{0}}}{\Delta x}$

$= \lim_{\Delta x \to 0}\frac{e^{x_{0}+\Delta x}-e^{x_{0}}}{\Delta x} = e^{x_{0}}.\lim_{\Delta x \to 0}\frac{e^{\Delta x}-1}{\Delta x}$

Đặt $e^{\Delta x} = n + 1$. Suy ra $\Delta x =ln(n+1)$. Khi $\Delta x \to 0$ thì $n \to 0$

Ta có: $y'(x_{0}) = e^{x_{0}}.\lim_{\Delta x \to 0}\frac{e^{\Delta x}-1}{\Delta x} = e^{x_{0}}.\lim_{n \to 0}\frac{n}{ln(n+1)}$

$= e^{x_{0}}.\lim_{n \to 0}\frac{1}{\frac{1}{n}.ln(n+1)} = e^{x_{0}}.\lim_{n \to 0}\frac{1}{ln(n+1)^{\frac{1}{n}}}$

Mà $\lim_{n \to 0}(n+1)^{\frac{1}{n}} = e$

Suy ra  $y'(x_{0})  = e^{x_{0}}.\frac{1}{lne} = e^{x_{0}}$

b) Ta có y = lnx. Suy ra $x = e^{y}$

Đạo hàm 2 vế, ta có:

$x' = y'.(e^{y})'$

$\Leftrightarrow 1 = y'.e^{y}$

$\Leftrightarrow y' = \frac{1}{e^{y}}$

$\Leftrightarrow y' = \frac{1}{x}$