Giải bài 2.66 trang 133 - SBT Giải tích lớp 12
Tính đạo hàm của các hàm số sau :
\(\begin{align} & a)\,y=\dfrac{1}{{{\left( 2+3x \right)}^{2}}}; \\ & b)\,y=\sqrt[3]{{{\left( 3x-2 \right)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x\ne \dfrac{2}{3} \right); \\ & c)\,y=\dfrac{1}{\sqrt[3]{3x-7}}; \\ & d)\,y=3{{x}^{-3}}-{{\log }_{3}}x; \\ & e)\,y=\left( 3{{x}^{2}}-2 \right){{\log }_{2}}x; \\ & g)\,y=\ln \left( \cos x \right); \\ & h)\,y={{e}^{x}}\sin x; \\ & i)\,y=\dfrac{{{e}^{x}}+{{e}^{-x}}}{x} \\ \end{align} \)
Gợi ý:
\((u^n)'=n.u'.u^{n-1}\\ (\log_au)'=\dfrac{u'}{u\ln a}\\ (e^u)'=u'e^u\\ (\ln u)'=\dfrac{u'}{u}\)
\(\begin{aligned} & a) \\ & y'=-\dfrac{2.3\left( 2+3x \right)}{{{\left( 2+3x \right)}^{4}}}=-\dfrac{6}{{{\left( 2+3x \right)}^{3}}} \\ \end{aligned} \)
\(\begin{aligned} & b) \\ & y={{\left( 3x-2 \right)}^{\frac{2}{3}}} \\ & \Rightarrow y'=\dfrac{2}{3}.3.{{\left( 3x-2 \right)}^{\frac{2}{3}-1}}=2{{\left( 3x-2 \right)}^{\frac{-1}{3}}} \\ \end{aligned} \)
\(\begin{aligned} & c) \\ & y={{\left( 3x-7 \right)}^{\frac{-1}{3}}} \\ & \Rightarrow y'=-\dfrac{1}{3}.3.{{\left( 3x-7 \right)}^{\frac{-1}{3}-1}}=-{{\left( 3x-7 \right)}^{\frac{-4}{3}}} \\ \end{aligned} \)
\(\begin{aligned} & d) \\ & y'=3.\left( -3 \right).{{x}^{-3-1}}-\dfrac{1}{x\ln 3} \\ \end{aligned}=-9x^{-4}-\dfrac{1}{x\ln 3}\)
\(\begin{aligned} & e) \\ & y'=6x{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x\ln 2} \\ \end{aligned} \)
\(\begin{aligned} & g) \\ & y'=\dfrac{-\sin x}{\cos x}=-\tan x \\ \end{aligned} \)
\(h)\,y'={{e}^{x}}\sin x+{{e}^{x}}\cos x \)
\(i)\,y'=\dfrac{\left( {{e}^{x}}-{{e}^{-x}} \right)x-\left( {{e}^{x}}+{{e}^{-x}} \right)}{{{x}^{2}}}=\dfrac{\left( x-1 \right)(e^x-{{e}^{-x})}}{{{x}^{2}}} \)