Tìm đạo hàm của các hàm số sau Bài tập 2 trang 176 SGK Toán lớp 11 Đại số

Giải Toán 11 Ôn tập chương 5

Bài tập 2 trang 176 SGK Toán lớp 11 Đại số:

a) $y=2\sqrt{x}\sin x-\frac{\cos x}{x}$;

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

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

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

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

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

Lời giải:

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

$\begin{array}{l}{y}^{\text{'}}={\left(2\sqrt{x}\sin \text{x}-\frac{\cos x}{x}\right)}^{\text{'}}\\ =2{\left(\sqrt{x}\sin x\right)}^{\text{'}}-{\left(\frac{\cos x}{x}\right)}^{\text{'}}\\ =2\left.{\left(\sqrt{x}\right)}^{\text{'}}\sin x+\sqrt{x}.(\sin x{)}^{\text{'}}\right]-\frac{(\cos x{)}^{\text{'}}.x-x^{\text{'}}\cos x}{x^2}\\ =2\frac{1}{2\sqrt{x}}\sin x+2\sqrt{x}\cos x-\frac{-x\sin x-\cos x}{x^2}\\ =\frac{\sqrt{x}\sin x}{x}+2\sqrt{x}\cos x+\frac{x\sin x+\cos x}{x^2}\\ =\frac{x\sqrt{x}\sin x+2x^2\sqrt{x}\cos x+x\sin x+\cos x}{x^2}\end{array}$

b) $y=\frac{3\cos x}{2x+1}$

$\begin{array}{l}{y}^{\text{'}}=\frac{3\left(\cos x{)}^{\text{'}}\right(2x+1)-3\cos x(2x+1{)}^{\text{'}}}{{(2x+1)}^2}\\ =\frac{-3\sin x(2x+1)-2.3\cos x}{{(2x+1)}^2}\\ =\frac{-6x\sin x-3\sin x-6\cos x}{{(2x+1)}^2}\end{array}$

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

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(t^2+2\cos t\right)}^{\text{'}}\sin t-{\left(\sin t\right)}^{\text{'}}\left(t^2+2\cos t\right)}{{\sin }^{2}t}\\ =\frac{(2t-2\sin t)\sin t-\cos t\left(t^2+2\cos t\right)}{{\sin }^{2}t}\\ =\frac{2t\sin t-2{\sin }^{2}t-t^2\cos t-2{\cos }^{2}t}{{\sin }^{2}t}\\ =\frac{2t\sin t-t^2\cos t-2\left({\sin }^{2}t+{\cos }^{2}t\right)}{{\sin }^{2}t}\\ =\frac{2t\sin t-t^2\cos t-2}{{\sin }^{2}t}\end{array}$

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

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(2\cos \phi -\sin \phi \right)}^{\text{'}}\left(3\sin \phi +\cos \phi \right)-\left(2\cos \phi -\sin \phi \right){\left(3\sin \phi +\cos \phi \right)}^{\text{'}}}{{\left(3\sin \phi +\cos \phi \right)}^2}\\ =\frac{\left(-2\sin \phi -\cos \phi \right)\left(3\sin \phi +\cos \phi \right)-\left(3\cos \phi -\sin \phi \right)\left(2\cos \phi -\sin \phi \right)}{{\left(3\sin \phi +\cos \phi \right)}^2}\\ =\frac{-6{\sin }^2\phi -5\sin \phi \cos \phi -{\cos }^2\phi -6{\cos }^2\phi +5\sin \phi \cos \phi -{\sin }^2\phi }{{\left(3\sin \phi +\cos \phi \right)}^2}\\ =\frac{-7{\sin }^2\phi -7{\cos }^2\phi }{{\left(3\sin \phi +\cos \phi \right)}^2}\\ =\frac{-7\left({\sin }^2\phi +{\cos }^2\phi \right)}{{\left(3\sin \phi +\cos \phi \right)}^2}\\ =\frac{-7}{{\left(3\sin \phi +\cos \phi \right)}^2}\end{array}$

e) $y=\frac{\tan x}{\sin x+2}$

$\begin{array}{l}{y}^{\text{'}}=\frac{\left(\tan x{)}^{\text{'}}\right(\sin x+2)-\tan x(\sin x+2{)}^{\text{'}}}{{(\sin x+2)}^2}\\ =\frac{\frac{1}{{\cos }^{2}x}(\sin x+2)-\tan x\cos x}{{(\sin x+2)}^2}\\ =\frac{\frac{\sin x+2}{{\cos }^{2}x}-\frac{\sin x}{\cos x}\cdot \cos x}{{(\sin x+2)}^2}\\ =\frac{\sin x+2-\sin x{\cos }^{2}x}{{\cos }^{2}x{(\sin x+2)}^2}\\ =\frac{\sin x\left(1-{\cos }^{2}x\right)+2}{{\cos }^{2}x{(\sin x+2)}^2}\\ =\frac{\sin x.{\sin }^{2}x+2}{{\cos }^{2}x{(\sin x+2)}^2}\\ =\frac{{\sin }^{3}x+2}{{\cos }^{2}x{(\sin x+2)}^2}\end{array}$

f)

$\begin{array}{l}y=\frac{\cot x}{2\sqrt{x}-1}\\ y^{\text{'}}=\frac{(\cot x{)}^{\text{'}}.\left(2\sqrt{x}-1\right)-\cot x.{\left(2\sqrt{x}-1\right)}^{\text{'}}}{{\left(2\sqrt{x}-1\right)}^2}=\frac{\frac{-1}{{\sin }^{2}x}\left(2\sqrt{x}-1\right)-\cot x\cdot \frac{1}{\sqrt{x}}}{{\left(2\sqrt{x}-1\right)}^2}.\end{array}$