Nội dung bài giảng
Tính đạo hàm của các hàm số sau:
a) \(y = \frac{1}{{{{(2 + 3x)}^2}}}\)
b) \(y = \sqrt[3]{{{{(3x - 2)}^2}}}(x \ne \frac{2}{3})\)
c) \(y = \frac{1}{{\sqrt[3]{{3x - 7}}}}\)
d) \(y = 3{x^{ - 3}} - {\log _3}x\)
e) \(y = (3{x^2} - 2){\log _2}x\)
g) \(y = \ln (\cos x)\)
h) \(y = {e^x}\sin x\)
i) \(y = \frac{{{e^x} - {e^{ - x}}}}{x}\)
Hướng dẫn làm bài:
a) \(y' = - 6{(2 + 3x)^{ - 3}}\)
b)
\(y' = \left\{ {\begin{array}{*{20}{c}}
{2{{(3x - 2)}^{ - \frac{1}{3}}},\forall x > \frac{2}{3}}\\
{ - 2{{(2 - 3x)}^{ - \frac{1}{3}}},\forall x < \frac{2}{3}}
\end{array}} \right. = \frac{2}{{\sqrt[3]{{3x - 2}}}}(x \ne \frac{2}{3})\)
c) \(y' = - \frac{1}{{\sqrt[3]{{{{(3x - 7)}^4}}}}}\)
d) \(y' = - 9{x^{ - 4}} - \frac{1}{{x\ln 3}}\)
e) \(y' = 6x{\log _2}x + \frac{{3{x^2} - 2}}{{x\ln 2}}\)
g) \(y' = - \tan x\)
h) \(y' = {e^x}(\sin x + \cos x)\)
i) \(y' = \frac{{x({e^x} + {e^{ - x}}) - {e^x} + {e^{ - x}}}}{{{x^2}}}\).