Nội dung bài giảng
Bài 2. Tính đạo hàm của các hàm số sau
a) \(y = 2\sqrt x {\mathop{\rm sinx}\nolimits} - {{\cos x} \over x}\)
b) \(y = {{3\cos x} \over {2x + 1}}\)
c) \(y = {{{t^2} + 2\cot t} \over {\sin t}}\)
d) \(y = {{2\cos \varphi - \sin \varphi } \over {3\sin \varphi + \cos \varphi }}\)
e) \(y = {{\tan x} \over {\sin x + 2}}\)
f) \(y = {{\cot x} \over {2\sqrt x - 1}}\)
Trả lời:
a)
\(y' =\left (2\sqrt x {\mathop{\rm sinx}\nolimits} - {{\cos x} \over x}\right)'\)
\(\eqalign{
& = 2{1 \over {2\sqrt x }}\sin x + 2\sqrt x\cos x - {{ - x\sin x - \cos x} \over {{x^2}}} \cr
& = {{x\sqrt x \sin x + 2{x^2}\sqrt x\cos x + x\sin x + \cos x} \over {{x^2}}} \cr
& = {{x(\sqrt x + 1)\sin x + (2{x^2}\sqrt x + 1)cosx} \over {{x^2}}} \cr} \)
b)
\(\eqalign{
& y' =\left ({{3\cos x} \over {2x + 1}}\right)' = {{ - 3(2x + 1)\sin x - 2.3\cos x} \over {{{(2x + 1)}^2}}} \cr
& = {{ - 3(2x + 1)\sin x - 6\cos x} \over {{{(2x + 1)}^2}}} \cr} \)
c)
\(\eqalign{
& y' = \left ({{{t^2} + 2\cot t} \over {\sin t}}\right )' = {{(2t - 2\sin t)\sin t - \cos t({t^2} + 2\cos t)} \over {{{\sin }^2}t}} \cr
& = {{2t\sin t - 2{{\sin }^2}t - {t^2}\cos t - 2{{\cos }^2}t} \over {{{\sin }^2}t}} \cr
& = {{2t\sin t - {t^2}\cos t - 2({{\sin }^2}t + {{\cos }^2}t)} \over {{{\sin }^2}t}} = {{2t\sin t - {t^2}\cos t - 2} \over {{{\sin }^2}t}} \cr} \)
d)
\(\eqalign{
& y' = \left({{2\cos \varphi - \sin \varphi } \over {3\sin \varphi + \cos \varphi }}\right)' \cr
& = {{( - 2sin\varphi - \cos \varphi )(3sin\varphi + \cos \varphi ) - (3\cos \varphi - \sin \varphi )(2\cos \varphi - \sin \varphi )} \over {{{(3\sin \varphi + \cos \varphi )}^2}}} \cr
& = {{ - 7} \over {{{(3\sin \varphi + \cos \varphi )}^2}}} \cr} \)
e)
\(\eqalign{
& y' = \left({{\tan x} \over {\sin x + 2}}\right)' = {{{1 \over {{{\cos }^2}x}}(\sin x + 2) - \cos x\tan x} \over {{{(\sin x + 2)}^2}}} = {{{1 \over {{{\cos }^2}x}}(\sin x + 2) - \sin x} \over {{{(\sin x + 2)}^2}}} \cr
& = {{\sin x + 2 - \sin x{{\cos }^2}x} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} = {{\sin x(1 - {{\cos }^2}x) + 2} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} = {{{{\sin }^3}x + 2} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} \cr} \)
f)
\(\eqalign{
& y' = \left({{\cot x} \over {2\sqrt x - 1}}\right)' = {{(\cot x)'(2\sqrt x - 1) - \cot x(2\sqrt x - 1)'} \over {{{(2\sqrt x - 1)}^2}}} = {{{{ - 1} \over {{{\sin }^2}x}}(2\sqrt x - 1) - \cot x.{1 \over {\sqrt x }}} \over {{{(2\sqrt x - 1)}^2}}} \cr
& = {{{{1 - 2\sqrt x } \over {{{\sin }^2}x}} - {{\cot x} \over {\sqrt x }}} \over {{{(2\sqrt x - 1)}^2}}} \cr} \)