a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^x} - {e^{{x_0}}}}}{{x - {x_0}}}\)

Đặt \(x = {x_0} + \Delta x\). Ta có:

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0} + \Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.{e^{\Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.\left( {{e^{\Delta x}} - 1} \right)}}{{\Delta x}}\\ &  = \mathop {\lim }\limits_{\Delta x \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}} = {e^{{x_0}}}.1 = {e^{{x_0}}}\end{array}\)

Vậy \({\left( {{e^x}} \right)^\prime } = {e^x}\) trên \(\mathbb{R}\).

b) Với bất kì \({x_0} > 0\), ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln {\rm{x}} - \ln {{\rm{x}}_0}}}{{x - {x_0}}}\)

Đặt \(x = {x_0} + \Delta x\). Ta có:

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{x_0} + \Delta x} \right) - \ln {{\rm{x}}_0}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {\frac{{{x_0} + \Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}}\end{array}\)

Đặt \(\frac{{\Delta x}}{{{x_0}}} = t\). Lại có: \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}} = \frac{1}{{{x_0}}};\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\ln \left( {1 + t} \right)}}{t} = 1\)

Vậy \(f'\left( {{x_0}} \right) = \frac{1}{{{x_0}}}.1 = \frac{1}{{{x_0}}}\)

Vậy \({\left( {\ln x} \right)^\prime } = \frac{1}{x}\) trên khoảng \(\left( {0; + \infty } \right)\).

\(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)

Ta có : \(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x=\lim\limits_{x\rightarrow+\infty}\left(1-\frac{1}{1+x}\right)^x\)

Đặt \(-\frac{1}{1+x}=\frac{1}{t}\Rightarrow\begin{cases}x=-\left(1+t\right)\\x\rightarrow+\infty;t\rightarrow-\infty\end{cases}\)

\(\Rightarrow L=\lim\limits_{t\rightarrow-\infty}\left(1+\frac{1}{t}\right)^{-\left(1+t\right)}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)^{1+t}}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)\left(1+\frac{1}{t}\right)^t}=\frac{1}{1.e}=\frac{1}{e}\)

\(L=\lim\limits_{x\rightarrow0}\frac{e^x-e^{-x}}{\sin x}=\lim\limits_{x\rightarrow0}\frac{e^x-\frac{1}{e^x}}{\sin x}=\lim\limits_{x\rightarrow0}\frac{e^{2x}-1}{e^x\sin x}=\lim\limits_{x\rightarrow0}\frac{e^{2x}-1}{2x.\frac{\sin x}{2x}.e^x}\)

   \(=\lim\limits_{x\rightarrow0}\frac{e^{2x}-1}{2x}.\frac{1}{\frac{\sin x}{x}}.\frac{2}{e^x}=1.\frac{1}{1}.\frac{2}{1}=2\)

a) Với x > 0 bất kì và \(h = x - {x_0}\) ta có

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{x_0} + h} \right) - \ln {x_0}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}.{x_0}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}}} = \frac{1}{{{x_0}}}\end{array}\)

Vậy hàm số \(y = \ln x\) có đạo hàm là hàm số \(y' = \frac{1}{x}\)

b) Ta có \({\log _a}x = \frac{{\ln x}}{{\ln a}}\) nên \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)

\(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}=\lim\limits_{x\rightarrow+\infty}\left(1+\frac{3}{x-2}\right)^{2x-1}\)

Đặt \(\begin{cases}\frac{3}{x-2}=\frac{1}{t}\Rightarrow x=3t+2\\x\rightarrow+\infty;t\rightarrow+\infty\end{cases}\)

\(L=\lim\limits_{x\rightarrow+\infty}\left(1+\frac{1}{t}\right)^{6t+3}=\lim\limits_{x\rightarrow+\infty}\left\{\left[\left(1+\frac{1}{t}\right)^t\right]^6.\left(1+\frac{1}{t}\right)^3\right\}=e^6.1^3=e^6\)

\(L=\lim\limits_{x\rightarrow0}\frac{e^x-1}{\sqrt{x+1}-1}=\lim\limits_{x\rightarrow0}\frac{\left(e^x-1\right)\left(\sqrt{x+1}-1\right)}{x}=\lim\limits_{x\rightarrow0}\left[\frac{e^x-1}{x}.\left(\sqrt{x+1}-1\right)\right]=1.0=0\)

\(L=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{x^3.\frac{2}{x^2}}=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+x^3\right)}{x^3}.\frac{x^3}{2}\right]=1.0=0\)

\(L=\lim\limits_{x\rightarrow0}\frac{\ln x-1}{\tan x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\frac{\sin x}{\cos x}}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{2x.\frac{\sin x}{x}.\frac{1}{2\cos x}}\)

   \(=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+2x\right)}{2x}.\frac{1}{\frac{\sin x}{x}}.2\cos x\right]=1.\frac{1}{1}.2.1=2\)

a) Với x bất kì và \(h = x - {x_0}\), ta có:

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_0} + h}} - {e^{{x_0}}}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_o}}}\left( {{e^h} - 1} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = {e^{{x_0}}}\end{array}\)

Vậy hàm số \(y = {e^x}\)  có đạo hàm là hàm số \(y' = {e^x}\)

b) Ta có \({a^x} = {e^{x\ln a}}\,\)nên \(\left( {{a^x}} \right)' = \left( {{e^{x\ln a}}} \right)' = \left( {x\ln a} \right)'.{e^{x\ln a}} = {e^{x\ln a}}\ln a = {a^x}\ln a\)