Đạo hàm của hàm số y= 3sin 4x - 4cos 3x + 2019
A. 12cos 4x + 12sin 3x
B. 12cos 4x - 12sin 3x
C. 12sin 4x + 12cos 3x
D. 12cos 4x - 12sin 4x
K
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ND
Nguyễn Đức Trí
VIP
23 tháng 7 2023
\(tanx=\dfrac{1}{cotx}=\dfrac{1}{\sqrt[]{2}}=\dfrac{\sqrt[]{2}}{2}\left(tanx.cotx=1\right)\)
\(1+tan^2x=\dfrac{1}{cos^2x}\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}=\dfrac{1}{1+\dfrac{1}{2}}\)
\(\Rightarrow cos^2x=\dfrac{2}{3}\Rightarrow cosx=\sqrt[]{\dfrac{2}{3}}\)
\(tanx=\dfrac{sinx}{cosx}\Rightarrow sinx=tanx.cosx=\dfrac{1}{\sqrt[]{2}}.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{3}\)
\(P=\dfrac{3sinx-2cosx}{12sin^3x+4cos^3x}=\dfrac{3.\dfrac{\sqrt[]{3}}{3}-2.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)
\(=\dfrac{\sqrt[]{3}-\dfrac{2\sqrt[]{6}}{3}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)
31 tháng 8 2023
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: \(\left(\sqrt{4x^2-3x+1}\right)'=\dfrac{\left(4x^2-3x+1\right)'}{2\sqrt{4x^2-3x+1}}\)
\(=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)
\(y'=\left(\sqrt{4x^2-3x+1}\right)'-4'=\dfrac{8x-3}{2\sqrt{4x^2-3x+1}}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
4.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=cos2x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=cos2x\)
\(\Leftrightarrow1+1-sin^22x=2cos2x\)
\(\Leftrightarrow1+cos^22x=2cos2x\)
\(\Leftrightarrow\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow2x=k2\pi\)
\(\Rightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{1}{2}\)
\(\Leftrightarrow1-sin^22x=0\)
\(\Leftrightarrow cos^22x=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 5 2020
\(A=\frac{3sina-2cosa}{12sin^3a+4cos^3a}=\frac{\frac{3sina}{sin^3a}-\frac{2cosa}{sin^3a}}{12+\frac{4cos^3a}{sin^3a}}=\frac{3.\frac{1}{sin^2a}-2cota.\frac{1}{sin^2a}}{12+4cot^3a}\)
\(=\frac{3\left(1+cot^2a\right)-2cota\left(1+cot^2a\right)}{12+4cot^3a}=\frac{3\left(1+3^2\right)-2.3.\left(1+3^2\right)}{12+4.3^3}=...\)
17 tháng 8 2023
tham khảo:
a)y′=2\(^{3x-x^2}\).ln2.(3−2x)
b) y′\(\dfrac{4}{ln3}\).\(\dfrac{1}{4x+1}\).4=\(\dfrac{4}{\left(4x+1\right)ln3}\)
2
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.
\(y'=-7x^{-8}-\frac{(x^2-1)'}{x^2-1}=-7x^{-8}-\frac{2x}{x^2-1}\)
Đáp án A