Giải Hoạt động 9 trang 93 sgk Toán 11 tập 2 Kết nối

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

$=\lim_{h\to 0} \frac{e^{e\ln(x+h)}-e^{e\ln x}}{h}$

$=\lim_{h\to 0} \frac{e^{e\ln x}e^{e\ln(1+\frac{h}{x})}-e^{e\ln x}}{h}$

$=\lim_{h\to 0} \frac{e^{e\ln x}\left(e^{e\ln(1+\frac{h}{x})}-1\right)}{h}$

$=\lim_{h\to 0} \frac{e^{e\ln x}\left(e^{e\left(\frac{h}{x}+\mathcal{O}(h^2)\right)}-1\right)}{h}$

$=\lim_{h\to 0} \frac{e^{e\ln x}\left(e^{e\frac{h}{x}}-1\right)}{h} $

$=e^{e\ln x}\lim_{h\to 0} \frac{e^{e\frac{h}{x}}-1}{\frac{h}{x}}\cdot\frac{1}{x}$

$=x^e\lim_{h\to 0} \frac{e^{h/x}-1}{h/x}\cdot\frac{1}{x} \quad\text{(đặt $u=h/x$)}$

$=x^e\lim_{u\to 0} \frac{e^u-1}{u}\cdot\frac{1}{x}$

$=\frac{x^e}{x}$

b) $y'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}$

$=\lim_{h\to 0}\frac{e^{(x+h)\ln a}-e^{x\ln a}}{h}$

$=\lim_{h\to 0}\frac{e^{x\ln a}\left(e^{h\ln a}-1\right)}{h}$

$=a^x\lim_{h\to 0}\frac{e^{h\ln a}-1}{h}$

$=a^x\lim_{h\to 0}\frac{e^{h\ln a}-1}{h\ln a}\cdot\ln a$

$=a^x\cdot\ln a$