Giải câu 2 bài Ôn tập chương 5: Đạo hàm

a) \(y = 2\sqrt x {\mathop{\rm sinx}\nolimits}  - {{\cos x} \over x}\)

\(y' =\left (2\sqrt x {\mathop{\rm sinx}\nolimits}  - {{\cos x} \over x}\right)'\)

\(=2.(\sqrt{x}sin\,x)'-\left ( \frac{cos\, x}{x} \right )'\)

\(= 2{1 \over {2\sqrt x }}\sin x + 2\sqrt x\cos x - {{ - x\sin x - \cos x} \over {{x^2}}} \)

\(= {{x\sqrt x \sin x + 2{x^2}\sqrt x\cos x + x\sin x + \cos x} \over {{x^2}}} \)

\(= {{x(\sqrt x + 1)\sin x + (2{x^2}\sqrt x + 1)cosx} \over {{x^2}}} \)

b) \(y = {{3\cos x} \over {2x + 1}}\)

\(y' =\left ({{3\cos x} \over {2x + 1}}\right)' = {{ - 3(2x + 1)\sin x - 2.3\cos x} \over {{{(2x + 1)}^2}}}\)

\( = {{ - 3(2x + 1)\sin x - 6\cos x} \over {{{(2x + 1)}^2}}} \)

c) \(y = {{{t^2} + 2\cot t} \over {\sin t}}\)

\(y' = \left ({{{t^2} + 2\cos t} \over {\sin t}}\right )' \)

\(=\frac{(t^2+2cos\,t)'.sin\,t-(t^2+2cos\,t)(sin\,t)'}{sin^2t}\)

\(= {{(2t - 2\sin t)\sin t - \cos t({t^2} + 2\cos t)} \over {{{\sin }^2}t}} \)

\(= {{2t\sin t - 2{{\sin }^2}t - {t^2}\cos t - 2{{\cos }^2}t} \over {{{\sin }^2}t}} \)

\(= {{2t\sin t - {t^2}\cos t - 2({{\sin }^2}t + {{\cos }^2}t)} \over {{{\sin }^2}t}} \)

\(= {{2t\sin t - {t^2}\cos t - 2} \over {{{\sin }^2}t}} \)

d) \(y = {{2\cos \varphi  - \sin \varphi } \over {3\sin \varphi  + \cos \varphi }}\)

\(y' = \left({{2\cos \varphi - \sin \varphi } \over {3\sin \varphi + \cos \varphi }}\right)' \)

\(=\frac{(2cos\,\varphi -sin\,\varphi )'(3sin\,\varphi +cos\,\varphi )-(2cos\,\varphi -sin\,\varphi )(3sin\,\varphi +cos\,\varphi )'}{(3sin\,\varphi +cos\,\varphi )^2}\)

\(= {{( - 2sin\varphi - \cos \varphi )(3sin\varphi + \cos \varphi ) - (3\cos \varphi - \sin \varphi )(2\cos \varphi - \sin \varphi )} \over {{{(3\sin \varphi + \cos \varphi )}^2}}} \)

\(=\frac{-6sin^3 \varphi -2sin\,\varphi \,cos\,\varphi -3sin\,\varphi \,cos\,\varphi -cos^2 \varphi-(6cos^3 \varphi -2sin\, \varphi \,cos\, \varphi -3sin\, \varphi \,cos \,\varphi +sin^2 \varphi ) }{(3\sin \,\varphi + \cos \,\varphi )^2}\)

\( = {{ - 7} \over {{{(3\sin \varphi + \cos \varphi )}^2}}} \)

e) \(y = {{\tan x} \over {\sin x + 2}}\)

\(y' = \left({{\tan x} \over {\sin x + 2}}\right)' \)

\(= {{{1 \over {{{\cos }^2}x}}(\sin x + 2) - \cos x\tan x} \over {{{(\sin x + 2)}^2}}} \)

\(= {{{1 \over {{{\cos }^2}x}}(\sin x + 2) - \sin x} \over {{{(\sin x + 2)}^2}}} \)

\(= {{\sin x + 2 - \sin x{{\cos }^2}x} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} \)

\(= {{\sin x(1 - {{\cos }^2}x) + 2} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} \)

\(= {{{{\sin }^3}x + 2} \over {{{\cos }^2}x{{(\sin x + 2)}^2}}} \)

f) \(y = {{\cot x} \over {2\sqrt x  - 1}}\)

\(y' = \left({{\cot x} \over {2\sqrt x - 1}}\right)' \)

\(= {{(\cot x)'(2\sqrt x - 1) - \cot x(2\sqrt x - 1)'} \over {{{(2\sqrt x - 1)}^2}}} \)

\(= {{{{ - 1} \over {{{\sin }^2}x}}(2\sqrt x - 1) - \cot x.{1 \over {\sqrt x }}} \over {{{(2\sqrt x - 1)}^2}}} \)

\(= {{{{1 - 2\sqrt x } \over {{{\sin }^2}x}} - {{\cot x} \over {\sqrt x }}} \over {{{(2\sqrt x - 1)}^2}}} \)