Bài tập 29 trang 108 sgk Toán 11 tập 2 KNTT: Tính đạo hàm của các hàm số sau:
a) $y=3x^{2}-2 \sqrt{x}$
b) $y=\sqrt{1+2x-x^{2}}$
c) $y=tan \frac{x}{2}- cot \frac{x}{2}$
d) $y=e^{ex}+lnx{2}$
Bài Làm:
a) $y=3x^{2}-2 \sqrt{x}$
$y' = (3x^2)' - (2\sqrt{x})'$
$y' = 6x - \frac{1}{\sqrt{x}}$
b) $y=\sqrt{1+2x-x^{2}}$
$y' = \left(\sqrt{1+2x-x^2}\right)'$
$y' = \left(\sqrt{u}\right)' = \frac{1}{2\sqrt{u}}(u)'$
$y' = \frac{1}{2\sqrt{1+2x-x^2}}(1+2x-x^2)'$
$(1+2x-x^2)' = 2 - 2x$
$y' = \frac{1}{2\sqrt{1+2x-x^2}}(2 - 2x)$
$y' = \frac{2 - 2x}{2\sqrt{1+2x-x^2}}$
c) $y=tan \frac{x}{2}- cot \frac{x}{2}$
$y' = \left(\tan\frac{x}{2} - \cot\frac{x}{2}\right)'$$y' = \left(\tan\frac{x}{2} - \cot\frac{x}{2}\right)'$
$y' = \left(\tan\frac{x}{2}\right)' - \left(\cot\frac{x}{2}\right)'$
$y' = \frac{1}{\cos^2\frac{x}{2}}\cdot\frac{1}{2} - \frac{-1}{\sin^2\frac{x}{2}}\cdot\frac{1}{2}$
$y' = \frac{1}{2\cos^2\frac{x}{2}} + \frac{1}{2\sin^2\frac{x}{2}}$
d) $y = e^{ex} + \ln(x^2)$
$y' = (e^{ex} + \ln(x^2))'$
$y' = (e^{ex})' + (\ln(x^2))'$
$(e^{ex})' = e^{ex}\cdot(ex)'$
$(e^{ex})' = e^{ex}\cdot(e)'x + e^{ex}\cdot x'$
$(e^{ex})' = e^{ex}\cdot(e)\cdot x + e^{ex}\cdot 1$
$(e^{ex})' = e^{ex}\cdot(ex + 1)$
$(\ln(x^2))' = \frac{1}{x^2}\cdot(x^2)'$
$(\ln(x^2))' = \frac{1}{x^2}\cdot(2x)$
$(\ln(x^2))' = \frac{2x}{x^2}$
$y' = e^{ex}\cdot(ex + 1) + \frac{2x}{x^2}$
$y' = e^{ex}\cdot(ex + 1) + \frac{2}{x}$
