Tính đạo hàm f’(x) của hàm số f x = log 2 3 x - 1 với x >1/3
A. 3ln2/(3x-1).
B. 1/(3x-1)ln2.
C. 3/(3x-1).
D. 3/(3x-1)ln2.
K
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NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 11 2021
a.
\(y'=\dfrac{\left(1+\sqrt{3x-1}\right)'}{1+\sqrt{3x-1}}=\dfrac{3}{2\left(1+\sqrt{3x-1}\right)\sqrt{3x-1}}\)
b.
\(y'=\dfrac{\left(2sin^2x-1\right)'}{\left(2sin^2x-1\right).ln10}=\dfrac{2sin2x}{\left(2sin^2x-1\right)ln10}\)
c.
\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)
20 tháng 8 2023
a: \(y'=\left(x^2+3x-1\right)'\cdot e^x+\left(x^2+3x-1\right)\cdot\left(e^x\right)'\)
\(=e^x\left(2x+3\right)+\left(x^2+3x-1\right)\cdot e^x\)
\(=e^x\left(x^2+5x+2\right)\)
b: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)
\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)
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: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
ND
Nguyễn Đức Trí
VIP
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!
PT
1
22 tháng 2 2022
Bài 1:
a: f(0)=1
f(2)=-3x2+1=-6+1=-5
f(-2)=-3x2+1=-5
f(-1/2)=-3x1/2+1=-3/2+1=-1/2
b: f(x)=-3
=>-3|x|+1=-3
=>-3|x|=-4
=>|x|=4/3
=>x=4/3 hoặc x=-4/3
21 tháng 10 2023
2: ĐKXĐ: x<>1
\(f'\left(x\right)=\dfrac{\left(x^2-3x+3\right)'\left(x-1\right)-\left(x^2-3x+3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)
\(=\dfrac{\left(2x-3\right)\left(x-1\right)-\left(x^2-3x+3\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2-5x+3-x^2+3x-3}{\left(x-1\right)^2}=\dfrac{x^2-2x}{\left(x-1\right)^2}\)
f'(x)=0
=>x^2-2x=0
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
1:
\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)
=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)
f'(x)=0
=>\(\left(x-2\sqrt{2}\right)^2=0\)
=>\(x-2\sqrt{2}=0\)
=>\(x=2\sqrt{2}\)
14 tháng 8 2023
a: \(y'=\left(sin3x\right)'+\left(sin^2x\right)'=3\cdot cos3x+sin\left(x+pi\right)\)
b: \(y'=\left(log_2\left(2x+1\right)\right)'+\left(3^{-2x+1}\right)'\)
\(=\dfrac{2}{\left(2n+1\right)\cdot ln2}-2\cdot3^{-2x+1}\cdot ln3\)
Đáp án D