Tính đạo hàm của các hàm số sau
LG a
(y = {{{x^3}} over 3} – {{{x^2}} over 2} + x – 5)
Phương pháp giải:
Sử dụng bảng đạo hàm cơ bản và các quy tắc tính đạo hàm của tích, thương.
Lời giải chi tiết:
(begin{array}{l}y’ = left( {dfrac{{{x^3}}}{3}} right)’ – left( {dfrac{{{x^2}}}{2}} right)’ + left( x right)’ – left( 5 right)’\ = dfrac{{3{x^2}}}{3} – dfrac{{2x}}{2} + 1\ = {x^2} – x + 1end{array})
LG b
(displaystyle y = {2 over x} – {4 over {{x^2}}} + {5 over {{x^3}}} – {6 over {7{x^4}}})
Lời giải chi tiết:
(begin{array}{l}y’ = left( {dfrac{2}{x}} right)’ – left( {dfrac{4}{{{x^2}}}} right)’ + left( {dfrac{5}{{{x^3}}}} right)’ – left( {dfrac{6}{{7{x^4}}}} right)\ = – dfrac{2}{{{x^2}}} – dfrac{{ – 4.left( {{x^2}} right)’}}{{{x^4}}} + dfrac{{ – 5left( {{x^3}} right)’}}{{{x^6}}} – dfrac{{ – 6left( {{x^4}} right)’}}{{7{x^8}}}\ =- dfrac{2}{{{x^2}}} + dfrac{{4.2x}}{{{x^4}}} – dfrac{{5.3{x^2}}}{{{x^6}}} + dfrac{{6.4{x^3}}}{{7{x^8}}}\= – dfrac{2}{{{x^2}}} + dfrac{8}{{{x^3}}} – dfrac{{15}}{{{x^4}}} + dfrac{{24}}{{7{x^5}}}\end{array})
LG c
(displaystyle y = {{3{x^2} – 6x + 7} over {4x}})
Lời giải chi tiết:
(begin{array}{l}y’ = dfrac{{left( {3{x^2} – 6x + 7} right)’.4x – left( {3{x^2} – 6x + 7} right).left( {4x} right)’}}{{{{left( {4x} right)}^2}}}\= dfrac{{left( {6x – 6} right).4x – 4left( {3{x^2} – 6x + 7} right)}}{{16{x^2}}}\ = dfrac{{24{x^2} – 24x – 12{x^2} + 24x – 28}}{{16{x^2}}}\= dfrac{{12{x^2} – 28}}{{16{x^2}}} = dfrac{{3{x^2} – 7}}{{4{x^2}}}\end{array})
Cách khác:
(begin{array}{l}y = dfrac{3}{4}x – dfrac{3}{2} + dfrac{7}{{4x}}\y’ = left( {dfrac{3}{4}x} right)’ – left( {dfrac{3}{2}} right)’ + left( {dfrac{7}{{4x}}} right)’\ = dfrac{3}{4} – 0 – dfrac{7}{{4{x^2}}}\ = dfrac{{3{x^2} – 7}}{{4{x^2}}}end{array})
LG d
(displaystyle y = ({2 over x} + 3x)(sqrt x – 1))
Lời giải chi tiết:
(begin{array}{l}y’ = left( {dfrac{2}{x} + 3x} right)’left( {sqrt x – 1} right) + left( {dfrac{2}{x} + 3x} right)left( {sqrt x – 1} right)’\= left( { – dfrac{2}{{{x^2}}} + 3} right)left( {sqrt x – 1} right) + left( {dfrac{2}{x} + 3x} right).dfrac{1}{{2sqrt x }}\= dfrac{{ – 2}}{{xsqrt x }} + dfrac{2}{{{x^2}}} + 3sqrt x – 3 + dfrac{1}{{xsqrt x }} + dfrac{3}{2}sqrt x \,,,,,,,,,,,, = dfrac{{ – 1}}{{xsqrt x }} + dfrac{2}{{{x^2}}} + dfrac{{9sqrt x }}{2} – 3\end{array})
LG e
(displaystyle y = {{1 + sqrt x } over {1 – sqrt x }})
Lời giải chi tiết:
(begin{array}{l}y’ = dfrac{{left( {1 + sqrt x } right)’left( {1 – sqrt x } right) – left( {1 + sqrt x } right)left( {1 – sqrt x } right)’}}{{{{left( {1 – sqrt x } right)}^2}}}\ =dfrac{{dfrac{1}{{2sqrt x }}left( {1 – sqrt x } right) + dfrac{1}{{2sqrt x }}left( {1 + sqrt x } right)}}{{{{left( {1 – sqrt x } right)}^2}}}\= dfrac{1}{{sqrt x {{left( {1 – sqrt x } right)}^2}}}\end{array})
LG f
(displaystyle y = {{ – {x^2} + 7x + 5} over {{x^2} – 3x}})
Lời giải chi tiết:
(begin{array}{l}y’ = dfrac{{left( { – {x^2} + 7x + 5} right)’left( {{x^2} – 3x} right) – left( { – {x^2} + 7x + 5} right)left( {{x^2} – 3x} right)’}}{{{{left( {{x^2} – 3x} right)}^2}}}\= dfrac{{left( { – 2x + 7} right)left( {{x^2} – 3x} right) – left( {2x – 3} right)left( { – {x^2} + 7x + 5} right)}}{{{{left( {{x^2} – 3x} right)}^2}}}\= dfrac{{ – 2{x^3} + 13{x^2} – 21x + 2{x^3} – 17{x^2} + 11x + 15}}{{{{left( {{x^2} – 3x} right)}^2}}}\= dfrac{{ – 4{x^2} – 10x + 15}}{{{{left( {{x^2} – 3x} right)}^2}}}end{array})
