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ánh tuyết nguyễn
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Nguyễn Kiều Anh
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Nguyễn Việt Lâm
30 tháng 7 2021 lúc 22:51

1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)

2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)

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

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títtt
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Nguyễn Đức Trí
15 tháng 9 2023 lúc 19:47

1) \(f\left(x\right)=2x-5\)

\(f'\left(x\right)=2\)

\(\Rightarrow f'\left(4\right)=2\)

2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

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

3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

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

\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)

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Nguyễn Lê Phước Thịnh
15 tháng 9 2023 lúc 19:42

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títtt
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Nguyễn Lê Phước Thịnh
31 tháng 8 2023 lúc 21:12

a: \(y'=\left(x^2\right)'+\left(3x\right)'-\left(6x^6\right)'+\left(\dfrac{2x-3}{x-1}\right)'\)

\(=2x+3-6\cdot6x^5+\dfrac{\left(2x-3\right)'\left(x-1\right)-\left(2x-3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=-36x^5+2x+3+\dfrac{2\left(x-1\right)-2x+3}{\left(x-1\right)^2}\)

\(=-36x^5+2x+3+\dfrac{1}{\left(x-1\right)^2}\)

b: \(\left(\sqrt{2x^2-3x+1}\right)'=\dfrac{\left(2x^2-3x+1\right)'}{2\sqrt{2x^2-3x+1}}\)

\(=\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

\(y'=3\cdot2x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

\(=6x-4+\dfrac{4x-3}{2\sqrt{2x^2-3x+1}}\)

c: \(\left(\sqrt{4x^2-3x+1}\right)'=\dfrac{\left(4x^2-3x+1\right)'}{2\sqrt{4x^2-3x+1}}\)

\(=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)

\(y'=\left(\sqrt{4x^2-3x+1}\right)'-4'=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)

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nanako
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Hoàng Tử Hà
6 tháng 4 2021 lúc 13:54

a/ \(y=\left(x^3-3x\right)^{\dfrac{3}{2}}\Rightarrow y'=\dfrac{3}{2}\left(x^3-3x\right)^{\dfrac{1}{2}}\left(x^3-3x\right)'=\dfrac{3}{2}\left(3x^2-3\right)\sqrt{x^3-3x}\)

b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/ 

\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)

d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)

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nanako
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Hoàng Tử Hà
8 tháng 4 2021 lúc 20:55

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

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

b/ \(y'=\dfrac{\left(4x^3+2x-3\right)'\left(\sqrt{x^2+2}\right)-\left(\sqrt{x^2+2}\right)'\left(4x^3+2x-3\right)}{x^2+2}\)

\(y'=\dfrac{\left(12x^2+2\right)\sqrt{x^2+2}-\dfrac{x}{\sqrt{x^2+2}}\left(4x^3+2x-3\right)}{x^2+2}\) (ban tu rut gon nhe)

c/ \(y'=\dfrac{\left(x^3+x+1\right)'\left(x^3+x+1\right)}{\left|x^3+x+1\right|}=\dfrac{\left(3x^2+1\right)\left(x^3+x+1\right)}{\left|x^3+x+1\right|}\) 

d/ \(y'=\dfrac{3x^2-24x^3}{2\sqrt{x^3-6x^4+7}}\)

e/ \(y'=\dfrac{\left(x^5+1\right)'\left(2-\sqrt{x^2+3}\right)-\left(x^5+1\right)\left(2-\sqrt{x^2+3}\right)'}{\left(2-\sqrt{x^2+3}\right)^2}\)

\(y'=\dfrac{5x^4\left(2-\sqrt{x^2+3}\right)+\left(x^5+1\right)\dfrac{x}{\sqrt{x^2+3}}}{\left(2-\sqrt{x^2+3}\right)^2}\)

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títtt
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Nguyễn Đức Trí
17 tháng 9 2023 lúc 22:03

1) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

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

2) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1\left(x+9\right)}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{-6}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(1\right)=\dfrac{-6}{\left(1+3\right)^2}+\dfrac{2}{\sqrt[]{1}}=-\dfrac{3}{8}+2=\dfrac{13}{8}\)

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Sách Giáo Khoa
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Quang Duy
4 tháng 4 2017 lúc 19:31

Giải bài 1 trang 176 sgk Đại Số 11 | Để học tốt Toán 11

Giải bài 1 trang 176 sgk Đại Số 11 | Để học tốt Toán 11

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Nguyễn Phương Thùy
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Nguyễn Việt Lâm
10 tháng 9 2021 lúc 22:00

a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

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

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)

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SukhoiSu-35
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Bùi Nguyên Khải
17 tháng 8 2023 lúc 12:10

tham khảo:

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

\(y'=3x^2-6x+2\)

b)\(\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}\)

\(\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}\)

\(\dfrac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f'\left(x\right)+g'\left(x\right)\)

\(\dfrac{d}{dx}\left(cf\left(x\right)\right)=cf'\left(x\right)\)

\(y'=\dfrac{d}{dx}\left(x^2\right)-\dfrac{d}{dx}\left(4\sqrt{x}\right)+\dfrac{d}{dx}\left(3\right)\)

\(y'=2x-2\sqrt{x}\)

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