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25 tháng 11 2023

a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)

\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)

b: \(y=\left(3x+1\right)^{\Omega}\)

=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)

=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)

c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)

\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)

\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)

d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)

\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)

\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)

\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)

e: \(y=3^{x^2}\)

=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)

f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)

h: \(y=\left(x+1\right)\cdot e^{cosx}\)

=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)

=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)

\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)

25 tháng 11 2023

a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)

b) \(y=\left(3x+1\right)^{\pi}\)

\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)

c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)

d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)

e) \(y=3^{x^2}\)

\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)

f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)

Các bài còn lại bạn tự làm nhé!

22 tháng 11 2023

d: ĐKXĐ: \(x^2-1< >0\)

=>\(x^2\ne1\)

=>\(x\notin\left\{1;-1\right\}\)

Vậy: TXĐ là D=R\{1;-1}

b: ĐKXĐ: \(2-x^2>0\)

=>\(x^2< 2\)

=>\(-\sqrt{2}< x< \sqrt{2}\)

Vậy: TXĐ là \(D=\left(-\sqrt{2};\sqrt{2}\right)\)

a: ĐKXĐ: \(x-1>0\)

=>x>1

Vậy: TXĐ là \(D=\left(1;+\infty\right)\)

c: ĐKXĐ: \(x^2+x-6>0\)

=>\(x^2+3x-2x-6>0\)

=>\(\left(x+3\right)\left(x-2\right)>0\)

TH1: \(\left\{{}\begin{matrix}x+3>0\\x-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>2\\x>-3\end{matrix}\right.\)

=>x>2

TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -3\\x< 2\end{matrix}\right.\)

=>x<-3

Vậy: TXĐ là \(D=\left(2;+\infty\right)\cup\left(-\infty;-3\right)\)

e: ĐKXĐ: \(x^2-2>0\)

=>\(x^2>2\)

=>\(\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

Vậy: TXĐ là \(D=\left(-\infty;-\sqrt{2}\right)\cup\left(\sqrt{2};+\infty\right)\)

f: ĐKXĐ: \(\sqrt{x-1}>0\)

=>x-1>0

=>x>1

Vậy: TXĐ là \(D=\left(1;+\infty\right)\)

g: ĐKXĐ: \(x^2+x-6>0\)

=>\(\left(x+3\right)\left(x-2\right)>0\)

=>\(\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)

Vậy: TXĐ là \(D=\left(2;+\infty\right)\cup\left(-\infty;-3\right)\)

18 tháng 11 2023

`a)TXĐ: R`

`b)TXĐ: R\\{0}`

`c)TXĐ: R\\{1}`

`d)TXĐ: (-oo;-1)uu(1;+oo)`

`e)TXĐ: (-oo;-1/2)uu(1/2;+oo)`

`f)TXĐ: (-oo;-\sqrt{2})uu(\sqrt{2};+oo)`

`h)TXĐ: (-oo;0) uu(2;+oo)`

`k)TXĐ: R\\{1/2}`

`l)ĐK: {(x^2-1 > 0),(x-2 > 0),(x-1 ne 0):}`

`<=>{([(x > 1),(x < -1):}),(x > 2),(x ne 1):}`

`<=>x > 2`

   `=>TXĐ: (2;+oo)`

18 tháng 11 2023

câu l) $x^2-1 > 0$ thì giải ra 2 nghiệm $x < -1, x > 1$ mới đúng chứ nhỉ?

18 tháng 11 2023

`a)TXĐ:R\\{1;1/3}`

`y'=[-4(6x-4)]/[(3x^2-4x+1)^5]`

`b)TXĐ:R`

`y'=2x. 3^[x^2-1] ln 3-e^[-x+1]`

`c)TXĐ: (4;+oo)`

`y'=[2x-4]/[x^2-4x]+2/[(2x-1).ln 3]`

`d)TXĐ:(0;+oo)`

`y'=ln x+2/[(x+1)^2].2^[[x-1]/[x+1]].ln 2`

`e)TXĐ:(-oo;-1)uu(1;+oo)`

`y'=-7x^[-8]-[2x]/[x^2-1]`

AH
Akai Haruma
Giáo viên
18 tháng 11 2023

Lời giải:
a.

$y'=-4(3x^2-4x+1)^{-5}(3x^2-4x+1)'$

$=-4(3x^2-4x+1)^{-5}(6x-4)$

$=-8(3x-2)(3x^2-4x+1)^{-5}$

b.

$y'=(3^{x^2-1})'+(e^{-x+1})'$

$=(x^2-1)'3^{x^2-1}\ln 3 + (-x+1)'e^{-x+1}$

$=2x.3^{x^2-1}.\ln 3 -e^{-x+1}$

c.

$y'=\frac{(x^2-4x)'}{x^2-4x}+\frac{(2x-1)'}{(2x-1)\ln 3}$

$=\frac{2x-4}{x^2-4x}+\frac{2}{(2x-1)\ln 3}$

d.

\(y'=(x\ln x)'+(2^{\frac{x-1}{x+1}})'=x(\ln x)'+x'\ln x+(\frac{x-1}{x+1})'.2^{\frac{x-1}{x+1}}\ln 2\)

\(=x.\frac{1}{x}+\ln x+\frac{2}{(x+1)^2}.2^{\frac{x-1}{x+1}}\ln 2\\ =1+\ln x+\frac{2^{\frac{2x}{x+1}}\ln 2}{(x+1)^2}\)

e.

\(y'=-7x^{-8}-\frac{(x^2-1)'}{x^2-1}=-7x^{-8}-\frac{2x}{x^2-1}\)

24 tháng 5 2017

Nguyên hàm, tích phân và ứng dụng

Nguyên hàm, tích phân và ứng dụng

NV
26 tháng 3 2021

Đáp án B

Hàm bậc nhất trên bậc nhất không có cực trị

10 tháng 6 2021

Ta có: 

\(y'=\left(3^{x+1}\right)'\)

    \(=3^{x+1}ln3\)

\(\Rightarrow A\)

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