Giải câu 2 bài Ôn tập chương 5: Đạo hàm

Giải câu 2 bài Ôn tập chương 5: Đạo hàm.

a) $y=2\sqrt{x}\mathrm{sinx}-\frac{\cos x}{x}$

$y^{\prime }={(2\sqrt{x}\mathrm{sinx}-\frac{\cos x}{x})}^{\prime }$

$=2.(\sqrt{x}sinx{)}^{\prime }-{\left(\frac{cosx}{x}\right)}^{\prime }$

$=2\frac{1}{2\sqrt{x}}\sin x+2\sqrt{x}\cos x-\frac{-x\sin x-\cos x}{x^2}$

$=\frac{x\sqrt{x}\sin x+2x^2\sqrt{x}\cos x+x\sin x+\cos x}{x^2}$

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

b) $y=\frac{3\cos x}{2x+1}$

$y^{\prime }={\left(\frac{3\cos x}{2x+1}\right)}^{\prime }=\frac{-3(2x+1)\sin x-2.3\cos x}{{(2x+1)}^2}$

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

c) $y=\frac{t^2+2\cot t}{\sin t}$

$y^{\prime }={\left(\frac{t^2+2\cos t}{\sin t}\right)}^{\prime }$

$=\frac{(t^2+2cost{)}^{\prime }.sint-(t^2+2cost)(sint{)}^{\prime }}{si{n}^{2}t}$

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

$=\frac{2t\sin t-2{\sin }^{2}t-t^2\cos t-2{\cos }^{2}t}{{\sin }^{2}t}$

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

$=\frac{2t\sin t-t^2\cos t-2}{{\sin }^{2}t}$

d) $y=\frac{2\cos \phi -\sin \phi }{3\sin \phi +\cos \phi }$

$y^{\prime }={\left(\frac{2\cos \phi -\sin \phi }{3\sin \phi +\cos \phi }\right)}^{\prime }$

$=\frac{(2cos\phi -sin\phi {)}^{\prime }(3sin\phi +cos\phi )-(2cos\phi -sin\phi )(3sin\phi +cos\phi {)}^{\prime }}{(3sin\phi +cos\phi {)}^2}$

$=\frac{(-2sin\phi -\cos \phi )(3sin\phi +\cos \phi )-(3\cos \phi -\sin \phi )(2\cos \phi -\sin \phi )}{{(3\sin \phi +\cos \phi )}^2}$

$=\frac{-6si{n}^3\phi -2sin\phi cos\phi -3sin\phi cos\phi -co{s}^2\phi -(6co{s}^3\phi -2sin\phi cos\phi -3sin\phi cos\phi +si{n}^2\phi )}{(3\sin \phi +\cos \phi {)}^2}$

$=\frac{-7}{{(3\sin \phi +\cos \phi )}^2}$

e) $y=\frac{\tan x}{\sin x+2}$

$y^{\prime }={\left(\frac{\tan x}{\sin x+2}\right)}^{\prime }$

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

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

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

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

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

f) $y=\frac{\cot x}{2\sqrt{x}-1}$

$y^{\prime }={\left(\frac{\cot x}{2\sqrt{x}-1}\right)}^{\prime }$

$=\frac{(\cot x{)}^{\prime }(2\sqrt{x}-1)-\cot x(2\sqrt{x}-1{)}^{\prime }}{{(2\sqrt{x}-1)}^2}$

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

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