Bài 1. Tính đạo hàm của các hàm số sau
a) \(y = {{{x^3}} \over 3} - {{{x^2}} \over 2} + x - 5\)
b) \(y = {2 \over x} - {4 \over {{x^2}}} + {5 \over {{x^3}}} - {6 \over {7{x^4}}}\)
c) \(y = {{3{x^2} - 6x + 7} \over {4x}}\)
d) \(y = ({2 \over x} + 3x)(\sqrt x - 1)\)
e) \(y = {{1 + \sqrt x } \over {1 - \sqrt x }}\)
f) \(y = {{ - {x^2} + 7x + 5} \over {{x^2} - 3x}}\)
a)
\(\eqalign{
y’ &= \left ({{{x^3}} \over 3} - {{{x^2}} \over 2} + x - 5\right )’ \cr
& = {x^2} - x + 1 \cr} \)
b) Ta có: \(y = 2{x^{ - 1}} - 4{x^{-2}} + 5{x^{ - 3}} - {6 \over 7}{x^{-4}}\)
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\(\eqalign{
\Rightarrow y’& = - 2{x^{ - 2}} + 8{x^{ - 3}} - 15{x^{ - 4}} + {{24} \over 7}{x^{ - 5}} \cr
& = {{ - 2} \over {{x^2}}} + {8 \over {{x^3}}} - {{15} \over {{x^4}}} + {{24} \over {7{x^5}}} \cr} \)
c)
\(\eqalign{
& y’ = \left({{3{x^2} - 6x + 7} \over {4x}}\right)’ = {{(3{x^2} - 6x + 7)’4x - (4x)'(3{x^2} - 6x + 7)} \over {16{x^2}}} \cr
& = {{(6x - 6)4x - 4(3{x^2} - 6x + 7)} \over {16{x^2}}} = {{3{x^2} - 7} \over {4{x^2}}} \cr} \)
d) \(y’ = \left[ {({2 \over x} + 3x)(\sqrt x - 1)} \right]’\)
\(\eqalign{
& = ( - {2 \over {{x^2}}} + 3)(\sqrt x - 1) + {1 \over {2\sqrt x }}.({2 \over x} + 3x) \cr
& = - {{2\sqrt x } \over {{x^2}}} + {2 \over {{x^2}}} + 3\sqrt x - 3 + {1 \over {x\sqrt x }} + {{3x} \over {2\sqrt x }} \cr
& = - {{2\sqrt x } \over {{x^2}}} + {2 \over {{x^2}}} + 3\sqrt x - 3 + {{\sqrt x } \over {{x^2}}} + {{3\sqrt x } \over 2} = {{9{x^2}\sqrt x - 6{x^2} - 2\sqrt x + 4} \over {2{x^2}}} \cr} \)
e)
\(\eqalign{
& y’ = ({{1 + \sqrt x } \over {1 - \sqrt x }})’ = {{{1 \over {2\sqrt x }}(1 - \sqrt x ) + {1 \over {2\sqrt x }}(1 + \sqrt x )} \over {{{(1 - \sqrt x )}^2}}} \cr
& = {{1 - \sqrt x + 1 + \sqrt x } \over {2\sqrt x {{(1 - \sqrt x )}^2}}} = {1 \over {\sqrt x {{(1 - \sqrt x )}^2}}} \cr} \)
f)
\(\eqalign{
& y’ = ({{ - {x^2} + 7x + 5} \over {{x^2} - 3x}})’ = {{( - 2x + 7)({x^2} - 3x) - (2x - 3)( - {x^2} + 7x + 5)} \over {{{({x^2} - 3x)}^2}}} \cr
& = {{ - 4{x^2} - 10x + 15} \over {{{({x^2} - 3x)}^2}}} \cr} \)