Đáp án:
\(\begin{array}{l}
1,\,\,\,\,y' = 14{x^6} + \dfrac{3}{{{x^2}}} + 2\\
2,\,\,\,\,f'\left( x \right) = - \dfrac{7}{{{{\left( {2 - 3x} \right)}^2}}}\\
3,\,\,\,\,y' = \cos x - 3\sin 3x\\
4,\,\,\,\,f'\left( x \right) = \dfrac{{ - 5x}}{{\sqrt {4 - 5{x^2}} }}\\
5,\,\,\,\,f'\left( 1 \right) = 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = 2{x^7} - \dfrac{3}{x} + 2x\\
\Rightarrow y' = \left( {2{x^7} - \dfrac{3}{x} + 2x} \right)'\\
= 2.7.{x^6} - 3.\dfrac{{ - 1}}{{{x^2}}} + 2\\
= 14{x^6} + \dfrac{3}{{{x^2}}} + 2\\
\Rightarrow y' = 14{x^6} + \dfrac{3}{{{x^2}}} + 2\\
2,\\
f\left( x \right) = \dfrac{{x - 3}}{{2 - 3x}}\\
\Rightarrow f'\left( x \right) = \left( {\dfrac{{x - 3}}{{2 - 3x}}} \right)'\\
= \dfrac{{\left( {x - 3} \right)'.\left( {2 - 3x} \right) - \left( {2 - 3x} \right)'.\left( {x - 3} \right)}}{{{{\left( {2 - 3x} \right)}^2}}}\\
= \dfrac{{1.\left( {2 - 3x} \right) - \left( { - 3} \right).\left( {x - 3} \right)}}{{{{\left( {2 - 3x} \right)}^2}}}\\
= \dfrac{{\left( {2 - 3x} \right) - \left( { - 3x + 9} \right)}}{{{{\left( {2 - 3x} \right)}^2}}}\\
= \dfrac{{2 - 3x + 3x - 9}}{{{{\left( {2 - 3x} \right)}^2}}}\\
= \dfrac{{ - 7}}{{{{\left( {2 - 3x} \right)}^2}}}\\
\Rightarrow f'\left( x \right) = - \dfrac{7}{{{{\left( {2 - 3x} \right)}^2}}}\\
3,\\
y = \sin x + \cos 3x\\
\Rightarrow y' = \left( {\sin x + \cos 3x} \right)'\\
= \cos x + \left( {3x} \right)'.\left( { - \sin 3x} \right)\\
= \cos x + 3.\left( { - \sin 3x} \right)\\
= \cos x - 3\sin 3x\\
\Rightarrow y' = \cos x - 3\sin 3x\\
4,\\
f\left( x \right) = \sqrt {4 - 5{x^2}} \\
\Rightarrow f'\left( x \right) = \left( {\sqrt {4 - 5{x^2}} } \right)'\\
= \dfrac{{\left( {4 - 5{x^2}} \right)'}}{{2\sqrt {4 - 5{x^2}} }}\\
= \dfrac{{ - 5.2x}}{{2\sqrt {4 - 5{x^2}} }}\\
= \dfrac{{ - 10x}}{{2\sqrt {4 - 5{x^2}} }}\\
= \dfrac{{ - 5x}}{{\sqrt {4 - 5{x^2}} }}\\
\Rightarrow f'\left( x \right) = \dfrac{{ - 5x}}{{\sqrt {4 - 5{x^2}} }}\\
5,\\
f\left( x \right) = \dfrac{1}{2}{x^3} + \sqrt x - 15\\
\Rightarrow f'\left( x \right) = \left( {\dfrac{1}{2}{x^3} + \sqrt x - 15} \right)'\\
= \dfrac{1}{2}.3{x^2} + \dfrac{1}{{2\sqrt x }}\\
= \dfrac{3}{2}{x^2} + \dfrac{1}{{2\sqrt x }}\\
f'\left( 1 \right) = \dfrac{3}{2}{.1^2} + \dfrac{1}{{2.\sqrt 1 }} = \dfrac{3}{2} + \dfrac{1}{2} = 2\\
\Rightarrow f'\left( 1 \right) = 2
\end{array}\)
