Đề bài
Tính đạo hàm của các hàm số sau:
a) \(y = \frac{{ - 3{x^2}}}{2} + \frac{2}{x} + \frac{{{x^3}}}{3}\);
b) \(y = \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\left( {{x^2} + 9} \right)\);
c) \(y = \frac{{{x^2} - 2x}}{{{x^2} + x + 1}};\)
d) \(y = \frac{{1 - 2x}}{{x + 1}}\);
e) \(y = x.{e^{2x + 1}}\);
g) \(y = \left( {2x + 3} \right){.3^{2x + 1}}\);
h) \(y = x{\ln ^2}x\);
i) \(y = {\log _2}\left( {{x^2} + 1} \right)\).
Sử dụng kiến thức về các quy tắc tính đạo hàm để tính:
a) \(\left( {u + v + {\rm{w}}} \right)' = u' + v' + {\rm{w}}',\left( {{x^\alpha }} \right)' = \alpha .{x^{\alpha - 1}}\left( {x > 0} \right)\)
b) \(\left( {u \pm v} \right)' = u' \pm v',\left( {{x^\alpha }} \right)' = \alpha .{x^{\alpha - 1}}\left( {x > 0} \right),c' = 0\) với c là hằng số.
c, d) \({\left( {\frac{u}{v}} \right)'} = \frac{{u'v - uv'}}{{{v^2}}}\left( {v = v\left( x \right) \ne 0} \right)\) , \(\left( {{x^\alpha }} \right)' = \alpha .{x^{\alpha - 1}}\left( {x > 0} \right)\)
e) \(\left( {uv} \right)' = u'v + uv',\left( {{e^{u\left( x \right)}}} \right)' = \left( {u\left( x \right)} \right)'{e^{u\left( x \right)}}\)
g) \(\left( {uv} \right)' = u'v + uv',\left( {{a^{u\left( x \right)}}} \right)' = \left( {u\left( x \right)} \right)'{a^{u\left( x \right)}}\ln a\left( {a > 0,a \ne 1} \right)\)
h) \(\left( {uv} \right)' = u'v + uv',\left( {\ln x} \right)' = \frac{1}{x}\left( {x > 0} \right),\left\{ {{{\left[ {u\left( x \right)} \right]}^\alpha }} \right\}' = \alpha {\left[ {u\left( x \right)} \right]^{\alpha - 1}}\left[ {u\left( x \right)} \right]'\)
i) \(\left( {{{\log }_a}u\left( x \right)} \right)' = \frac{{u'\left( x \right)}}{{u\left( x \right)\ln a}}\left( {u\left( x \right) > 0,a > 0,a \ne 1} \right)\)
Lời giải chi tiết
a) \(y' = {\left( {\frac{{ - 3{x^2}}}{2} + \frac{2}{x} + \frac{{{x^3}}}{3}} \right)'} = \frac{{ - 3.2x}}{2} - \frac{2}{{{x^2}}} + \frac{{3.{x^2}}}{3} = - 3x - \frac{2}{{{x^2}}} + {x^2}\);
b) Ta có: \(y = \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\left( {{x^2} + 9} \right) = \left( {{x^4} - 5{x^2} + 4} \right)\left( {{x^2} + 9} \right)\)
\( = {x^6} - 5{x^4} + 4{x^2} + 9{x^4} - 45{x^2} + 36 = {x^6} + 4{x^4} - 41{x^2} + 36\)
Do đó, \(y' = \left( {{x^6} + 4{x^4} - 41{x^2} + 36} \right)' = 6{x^5} + 16{x^3} - 82x\)
c) \(y' = {\left( {\frac{{{x^2} - 2x}}{{{x^2} + x + 1}}} \right)'} = \frac{{\left( {{x^2} - 2x} \right)'\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 2x} \right)\left( {{x^2} + x + 1} \right)'}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\)
\( = \frac{{\left( {2x - 2} \right)\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 2x} \right)\left( {2x + 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\)
\( = \frac{{2{x^3} + 2{x^2} + 2x - 2{x^2} - 2x - 2 - 2{x^3} - {x^2} + 4{x^2} + 2x}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} = \frac{{3{x^2} + 2x - 2}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\)
d) \(y' = {\left( {\frac{{1 - 2x}}{{x + 1}}} \right)'} = \frac{{\left( {1 - 2x} \right)'\left( {x + 1} \right) - \left( {1 - 2x} \right)\left( {x + 1} \right)'}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{ - 2\left( {x + 1} \right) - \left( {1 - 2x} \right)}}{{{{\left( {x + 1} \right)}^2}}}\)
\( = \frac{{ - 2x - 2 - 1 + 2x}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{ - 3}}{{{{\left( {x + 1} \right)}^2}}}\)
e) \(y' = \left( {x.{e^{2x + 1}}} \right)' = x'.{e^{2x + 1}} + x.\left( {{e^{2x + 1}}} \right)' = {e^{2x + 1}} + x.2.{e^{2x + 1}} = {e^{2x + 1}}\left( {2x + 1} \right)\);
g) \(y' = \left( {\left( {2x + 3} \right){{.3}^{2x + 1}}} \right)' = \left( {2x + 3} \right)'{.3^{2x + 1}} + \left( {2x + 3} \right).\left( {{3^{2x + 1}}} \right)'\)
\( = {2.3^{2x + 1}} + \left( {2x + 3} \right)\left( {2x + 1} \right)'{.3^{2x + 1}}\ln 3 = {2.3^{2x + 1}} + {2.3^{2x + 1}}\left( {2x + 3} \right)\ln 3\)\( = {2.3^{2x + 1}}\left[ {\left( {2x + 3} \right)\ln 3 + 1} \right]\)
h) \(y' = \left( {x{{\ln }^2}x} \right)' = x'{\ln ^2}x + x.\left( {{{\ln }^2}x} \right)' = {\ln ^2}x + 2x.\ln x.\frac{1}{x} = {\ln ^2}x + 2.\ln x\);
i) \(y' = \left[ {{{\log }_2}\left( {{x^2} + 1} \right)} \right]' = \frac{{\left( {{x^2} + 1} \right)'}}{{\left( {{x^2} + 1} \right)\ln 2}} = \frac{{2x}}{{\left( {{x^2} + 1} \right)\ln 2}}\)
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