Tìm đạo hàm của các hàm số sau :
LG a
\(y = {{a{x^3} + b{x^2} + c} \over {\left( {a + b} \right)x}}\) (a, b, c là các hằng số)
Lời giải chi tiết:
\(\eqalign{ & y' = \left[ {{a \over {a + b}}{x^2} + {b \over {a + b}}x + {c \over {\left( {a + b} \right)x}}} \right] ' \cr & = {{2a} \over {a + b}}x + {b \over {a + b}} - {c \over {\left( {a + b} \right){x^2}}} \cr & = {{2a{x^3} + b{x^2} - c} \over {\left( {a + b} \right){x^2}}} \cr} \)
LG b
\(y = {\left( {{x^3} - {1 \over {{x^3}}} + 3} \right)^4}\)
Lời giải chi tiết:
\[\begin{array}{l}
y' = 4{\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)^3}\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)'\\
= 4{\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)^3}\left( {3{x^2} - \frac{{ - \left( {{x^3}} \right)'}}{{{x^6}}}} \right)\\
= 4{\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)^3}\left( {3{x^2} + \frac{{3{x^2}}}{{{x^6}}}} \right)\\
= 4{\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)^3}\left( {3{x^2} + \frac{3}{{{x^4}}}} \right)\\
= 12{\left( {{x^3} - \frac{1}{{{x^3}}} + 3} \right)^3}\left( {{x^2} + \frac{1}{{{x^4}}}} \right)
\end{array}\]
LG c
\(y = {x^3}{\cos ^2}x\)
Lời giải chi tiết:
\[\begin{array}{l}
y' = \left( {{x^3}} \right)'{\cos ^2}x + {x^3}\left( {{{\cos }^2}x} \right)'\\
= 3{x^2}{\cos ^2}x + {x^3}.2\cos x\left( { - \sin x} \right)\\
= 3{x^2}{\cos ^2}x - {x^3}\sin 2x\\
= {x^2}\left( {3{{\cos }^2}x - x\sin 2x} \right)
\end{array}\]
LG d
\(y = \sin \sqrt {4 + {x^2}} \)
Lời giải chi tiết:
\[\begin{array}{l}
y' = \left( {\sqrt {4 + {x^2}} } \right)'.\cos \sqrt {4 + {x^2}} \\
= \frac{{\left( {4 + {x^2}} \right)'}}{{2\sqrt {4 + {x^2}} }}.\cos \sqrt {4 + {x^2}} \\
= \frac{{2x}}{{2\sqrt {4 + {x^2}} }}.\cos \sqrt {4 + {x^2}} \\
= \frac{{x\cos \sqrt {4 + {x^2}} }}{{\sqrt {4 + {x^2}} }}
\end{array}\]
LG e
\(y = \sqrt {1 + \tan \left( {x + {1 \over x}} \right)} \)
Lời giải chi tiết:
\[\begin{array}{l}
y' = \frac{{\left( {1 + \tan \left( {x + \frac{1}{x}} \right)} \right)'}}{{2\sqrt {1 + \tan \left( {x + \frac{1}{x}} \right)} }}\\
= \frac{{\left( {x + \frac{1}{x}} \right)'.\frac{1}{{{{\cos }^2}\left( {x + \frac{1}{x}} \right)}}}}{{2\sqrt {1 + \tan \left( {x + \frac{1}{x}} \right)} }}\\
= \frac{{\left( {1 - \frac{1}{{{x^2}}}} \right).\frac{1}{{{{\cos }^2}\left( {x + \frac{1}{x}} \right)}}}}{{2\sqrt {1 + \tan \left( {x + \frac{1}{x}} \right)} }}\\
= \frac{{{x^2} - 1}}{{{x^2}{{\cos }^2}\left( {x + \frac{1}{x}} \right)}}.\frac{1}{{2\sqrt {1 + \tan \left( {x + \frac{1}{x}} \right)} }}\\
= \frac{{{x^2} - 1}}{{2{x^2}{{\cos }^2}\left( {x + \frac{1}{x}} \right)\sqrt {1 + \tan \left( {x + \frac{1}{x}} \right)} }}
\end{array}\]
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