Đáp án:

Ta có: 

$dy=y^{\text{'}}dx={\left(x^2+2x\right)}^{\text{'}}dx$

Đáp án:

Ta có 

$dy=f\text{'}\left(x\right)dx=2(x-1)dx$

Lời giải

Ta có: 

$f^{\text{'}}\left(x\right)=6x-1\Rightarrow f^{\text{'}}\left(2\right)=11$

$\Rightarrow df\left(2\right)=f^{\text{'}}\left(2\right).\Delta x=11.0,1=1,1$

Lời giải

$f^{\text{'}}\left(x\right)=2\cos 2x\Rightarrow f^{\text{'}}\left(\frac{\pi }{3}\right)=-1$

$\Rightarrow df\left(\frac{\pi }{3}\right)=f^{\text{'}}\left(\frac{\pi }{3}\right).\Delta x=-0,01$

Lời giải

Ta có:

$\begin{array}{l}f\text{'}\left(x\right)=\frac{\text{[\hspace{0.17em}}{(1+\tan x)}^2\text{]}\text{'}}{{(1+\text{}\tan x)}^4}\\ =\frac{2(1+\tan x).(1+\text{}\tan x)\text{'}}{{(1+\text{}\tan x)}^4}\\ =\frac{2}{{(1+\text{}\tan x)}^3}.\frac{1}{{\text{cos}}^{2}x}\end{array}$

Vi phân của hàm số đã cho là:

$dy=\frac{2}{{\cos }^{2}x{\left(1+\tan x\right)}^3}dx$

Lời giải

Ta có:

$\begin{array}{l}y\text{'}=\frac{{\left(1+{\cos }^{2}2x\right)}^{\text{'}}}{2\sqrt{1+{\cos }^{2}2x}}\\ =\frac{2c\text{os2x. (}c\text{os2x)'}}{2\sqrt{1+{\cos }^{2}2x}}\\ =\frac{2c\text{os2x. (-2 sin2x)}}{2\sqrt{1+{\cos }^{2}2x}}\\ =\frac{-\sin 4x}{\sqrt{1+{\cos }^{2}2x}}\\ df\left(x\right)=\frac{-\sin 4x}{\sqrt{1+{\cos }^{2}2x}}dx\end{array}$

Lời giải

Ta có:

 $\begin{array}{l}{f}^{\text{'}}\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }\frac{x^2+x}{x}\\ =\text{}\underset{x\to 0^{+}}{\lim }(x+1)=0+\text{\hspace{0.17em}}1=1;\end{array}$ 

$f^{\text{'}}\left(0^{-}\right)=\underset{x\to 0^{-}}{\lim }\frac{x}{x}=1$ và $df\left(0\right)=dx$.

Lời giải

$y\text{'}=3x^2\Rightarrow y\text{'}\left(1\right)=3$

Vi phân của hàm số tại $x_0=1$ với số gia $\Delta x=0,01$.

$dy\left(1\right)=y\text{'}\left(1\right).\Delta x=3.0,01=0,03$

Lời giải

Ta có:  

$\begin{array}{l}{y}^{\text{'}}=\frac{(x+3)\text{'}.(1-2x)-(x+\text{}3).(1-2x)\text{'}}{{(1-2x)}^2}\\ =\frac{1(1-2x)-(x+\text{}3).(-2)}{{(1-2x)}^2}\\ =\frac{7}{{\left(1-2\text{x}\right)}^2}\end{array}$

$\Rightarrow y^{\text{'}}\left(3\right)=\frac{1}{7}\Rightarrow dy=\frac{1}{7}dx$

Lời giải

$\begin{array}{l}{y}^{\text{'}}=2\cos 3x.(c\text{os3x)'}\\ =2cos3x\left(-3\sin 3x\right)\\ =-3\sin 6\text{x}\\ \Rightarrow dy=-3\sin 6x\text{d}x\end{array}$

Lời giải

Ta có 

$\text{d}y={\left(x^3-5x+6\right)}^{\text{'}}\text{d}x=\left(3x^2-5\right)\text{d}x$

Lời giải

Ta có 

$\begin{array}{l}\text{d}y={\left(\frac{1}{3x^3}\right)}^{\text{'}}\text{d}x\\ =\frac{1}{3}.\frac{-\left(x^3\right)\text{'}}{{\left(x^3\right)}^2}dx\\ =\frac{1}{3}.\frac{-3x^2}{{\left(x^3\right)}^2}dx\\ =-\frac{1}{x^4}\text{d}x\end{array}$

Lời giải

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

$\begin{array}{l}=\frac{1}{2\sqrt{x\sqrt{x}}}.\frac{3}{2}\sqrt{x}=\frac{3}{4}.\frac{\sqrt{x}}{\sqrt{x\sqrt{x}}}\\ =\frac{3}{4}.\sqrt{\frac{x}{x\sqrt{x}}}\\ =\frac{3}{4}.\sqrt{\frac{1}{\sqrt{x}}}=\frac{3}{4}.\frac{1}{\sqrt{\sqrt{x}}}\end{array}$

Vi phân của hàm số đã cho là:

$dy=\frac{3}{4\sqrt{\sqrt{x}}}dx$

Lời giải

Ta có

$\begin{array}{l}\text{dy}={\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)}^{\text{'}}\text{dx\hspace{0.17em}}\\ \text{= }\frac{\frac{1}{2\sqrt{x}}.\frac{1}{{\cos }^2\sqrt{x}}.\sqrt{x}-\tan \sqrt{x}.\frac{1}{2\sqrt{x}}}{x}\text{dx }\end{array}$

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

Lời giải

Ta có:  

$dy=y^{\text{'}}dx\Rightarrow y^{\text{'}}=\frac{dy}{dx}$ mà

$\begin{array}{l}y\text{'}=1+\text{\hspace{0.17em}}\frac{(x^2+1)\text{'}}{2\sqrt{x^2+1}}\\ =1+\text{\hspace{0.17em}}\frac{x}{\sqrt{x^2+1}}\\ =\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\\ =\frac{y}{\sqrt{x^2+1}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{y}{\sqrt{x^2+1}}\\ \iff \sqrt{x^2+1}dy-ydx=0\end{array}$

Lời giải

Xét $f\left(x\right)=\sqrt[3]{x}$  thì $f^{\text{'}}\left(x\right)=\frac{1}{3.\sqrt[3]{x^2}}$.

Cho $x_0=27,\Delta x=-0,3$.

Theo công thức gần đúng 

$f\left(x_0+\Delta x\right)\approx f^{\text{'}}\left(x_0\right).\Delta x+f\left(x_0\right)$

$\Rightarrow \sqrt[3]{27,3}\approx \sqrt[3]{27}+\frac{1}{27}\left(-0,3\right)\approx 2,999.$

Lời giải

Xét $f\left(x\right)=\sin x$ với $29°=\frac{\pi }{6}-\frac{\pi }{180}\left(rad\right)$ .

Có $f^{\text{'}}\left(x\right)=\cos x$.

Chọn $x_0=\frac{\pi }{6}$, $\Delta x=-\frac{\pi }{180}$

$\Rightarrow \sin \left(\frac{\pi }{6}-\frac{\pi }{180}\right)\approx \sin \frac{\pi }{6}+\cos \left(\frac{\pi }{6}\right).\left(-\frac{\pi }{180}\right)\approx 0,4849$

Lời giải

$\begin{array}{l}{x}^{2}y+y^3=2\\ \Rightarrow d\left(x^{2}y\right)+d\left(y^3\right)=0\\ \iff 2xy\text{d}x+x^{2}dy+3y^{2}dy=0\end{array}$

 tại điểm $\left(1;1\right)$ ta có:

$\begin{array}{l}2dx+dy+3dy=0\\ \Rightarrow 4dy=-2dx\\ \Rightarrow \frac{dy}{dx}=-\frac{1}{2}=y^{\text{'}}\left(1\right)\end{array}$

Lời giải

$\begin{array}{l}{y}^{\text{'}}=c\text{os( sin x). (sinx})\text{'}\\ =\mathrm{cosx}.cos\left(\sin x\right)\end{array}$

$\Rightarrow dy=\cos x.cos\left(\sin x\right)dx$

Lời giải

Ta có :

Vi phân của hàm số đã cho là $dy=\frac{x^{2}dx}{{\left(x\cos x-\sin x\right)}^2}$

Lời giải

Đặt  $u=x^2\Rightarrow y=f\left(u\right)$

Từ $f^{\text{'}}\left(x\right)=x^2-1\Rightarrow f^{\text{'}}\left(u\right)=u^2-1$

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}\\ =f^{\text{'}}\left(u\right).\frac{du}{dx}\\ =\left(u^2-1\right)2x\\ =2x\left(x^4-1\right)\end{array}$

Lời giải

Ta có: $y\text{'}=2x\Rightarrow y\text{'}\left(1\right)=2$

Chọn $\Delta x=dx=0,01;x_0=1\Rightarrow y_0=1$ 

 $\begin{array}{l}dy=2.0,01=0,02\\ \Rightarrow \Delta y-dy=0,0001\end{array}$

Lời giải

$\cos y={\sin }^{2}x\Rightarrow -\sin y\text{d}y=\sin 2x\text{d}x$

$\Rightarrow \frac{dy}{dx}=\frac{\sin 2x}{-\sin y}=\frac{\sin 2x}{-\sqrt{1-{\cos }^{2}y}}$ (vì $\sin y>0$ )

$\Rightarrow \frac{dy}{dx}=y^{\text{'}}=\frac{-\sin \frac{\pi }{2}}{\sqrt{1-{\sin }^4\frac{\pi }{4}}}=-\frac{2}{\sqrt{3}}$

Lời giải

$\begin{array}{l}{y}^{\text{'}}=\frac{\left(-4x-2\right){\left(x^2+x+1\right)}^2-\left(-2x^2-2x\right)2\left(x^2+x+1\right)\left(2x+1\right)}{{\left(x^2+x+1\right)}^4}\\ =\frac{2(x^2+x+\text{}1).(2x+\text{}1).\left[-(x^2+x+1)-(-2x^2-2x)\right]}{{\left(x^2+x+1\right)}^4}\end{array}$

 $=\frac{2\left(2\text{x}+1\right)\left(x^2+x-2\right)}{{\left(x^2+x+1\right)}^3}$

Vi phân của hàm số đã cho là $dy=\frac{2\left(2x+1\right)\left(x^2+x-2\right)}{{\left(x^2+x+1\right)}^3}dx$

Lời giải

$\begin{array}{l}y\text{'}=c\text{os}\sqrt{x^2+\text{}\sin (c\text{osx)}}.\left(\sqrt{x^2+\text{}\sin (c\text{osx)}}\right)\text{'}\\ =c\text{os}\sqrt{x^2+\text{}\sin (c\text{osx)}}.\frac{1}{2\sqrt{x^2+\text{}\sin (c\text{osx)}}}.\left(x^2+\text{}\sin (c\text{osx)}\right)\text{'}\\ =\frac{c\text{os}\sqrt{x^2+\text{}\sin (c\text{osx)}}}{2\sqrt{x^2+\text{}\sin (c\text{osx)}}}.\text{[\hspace{0.17em}\hspace{0.17em}}2x+\text{}c\text{os(sinx}).(c\text{osx)']}\\ =\frac{c\text{os}\sqrt{x^2+\text{}\sin (c\text{osx)}}}{2\sqrt{x^2+\text{}\sin (c\text{osx)}}}.\text{[\hspace{0.17em}\hspace{0.17em}}2x-\text{\hspace{0.17em}\hspace{0.17em}}c\text{os(sinx}).\sin x\text{]}\end{array}$

Do đó, vi phân của hàm số là

$dy=\frac{c\text{os}\sqrt{x^2+\text{}\sin (c\text{osx)}}}{2\sqrt{x^2+\text{}\sin (c\text{osx)}}}.\text{[\hspace{0.17em}\hspace{0.17em}}2x-\text{\hspace{0.17em}\hspace{0.17em}}c\text{os(sinx}).\sin x\text{] dx}$

Lời giải

$\begin{array}{l}y\text{'}=\frac{1}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.\left\{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}\right\}\text{'}\\ =\frac{1}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.c\text{os[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}.\text{\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}\text{'}\\ =\frac{c\text{os[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.\left[2x+\text{\hspace{0.17em}\hspace{0.17em}}\frac{\text{[}\sin (2x-3)\text{]}\text{'}}{2\sqrt{\sin (2x-3)}}\right]\\ =\frac{c\text{os[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.\left[2x+\text{\hspace{0.17em}\hspace{0.17em}}\frac{\text{2.cos}(2x-3)}{2\sqrt{\sin (2x-3)}}\right]\\ =\frac{c\text{os[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.\left[2x+\text{\hspace{0.17em}\hspace{0.17em}}\frac{\text{cos}(2x-3)}{\sqrt{\sin (2x-3)}}\right]\end{array}$

Vi phân của hàm số đã cho là

$dy=\frac{c\text{os[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{\hspace{0.17em}]}}{2\sqrt{\sin \text{[\hspace{0.17em}}{x}^2+\text{}\sqrt{\sin (2x-3)}\text{]}}}.\left[2x+\text{\hspace{0.17em}\hspace{0.17em}}\frac{\text{cos}(2x-3)}{\sqrt{\sin (2x-3)}}\right]dx$