Tính đạo hàm của các hàm số sau:
a) \(y = {x^2} + 1;\)
b) \(y = kx + c\) (với k, c là các hằng số).
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NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
a. \(y'=\dfrac{-1}{\left(x-1\right)}\)
b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)
c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)
e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)
g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
2.
a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)
b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)
c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)
d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)
e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)
f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
17 tháng 8 2023
tham khảo:
a)\(y'=\dfrac{\left(2\right)\left(x+2\right)-\left(2x-1\right)\left(1\right)}{\left(x+2\right)^2}\)
\(y'=\dfrac{5}{\left(x+2\right)^2}\)
b)\(y'=\dfrac{\left(2\right)\left(x^2+1\right)-\left(2x\right)\left(2x\right)}{\left(x^2+1\right)^2}\)
\(y'=\dfrac{2\left(1-x^2\right)}{\left(x^2+1\right)^2}\)
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=\(\dfrac{4}{\left(4x+1\right)ln3}\)
B
Tính đạo hàm của các hàm số sau:
a) \(y = {x^3} - 3{x^2} + 2x + 1;\)
b) \(y = {x^2} - 4\sqrt x + 3.\)
1
17 tháng 8 2023
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)\(\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}\)
\(\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}\)
\(\dfrac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f'\left(x\right)+g'\left(x\right)\)
\(\dfrac{d}{dx}\left(cf\left(x\right)\right)=cf'\left(x\right)\)
\(y'=\dfrac{d}{dx}\left(x^2\right)-\dfrac{d}{dx}\left(4\sqrt{x}\right)+\dfrac{d}{dx}\left(3\right)\)
\(y'=2x-2\sqrt{x}\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023
\(a,y'=\left[\left(2x-3\right)^{10}\right]'\\ =10\left(2x-3\right)^9\left(2x-3\right)'\\ =20\left(2x-3\right)^9\\ b,y'=\left(\sqrt{1-x^2}\right)'\\ =\dfrac{\left(1-x^2\right)'}{2\sqrt{1-x^2}}\\ =-\dfrac{2x}{2\sqrt{1-x^2}}\\ =-\dfrac{x}{\sqrt{1-x^2}}\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023
\(a,y'=\left(\dfrac{\sqrt{x}}{x+1}\right)'\\ =\dfrac{\left(\sqrt{x}\right)'\left(x+1\right)-\sqrt{x}\left(x+1\right)}{\left(x+1\right)^2}\\ =\dfrac{\dfrac{x+1}{2\sqrt{x}}-\sqrt{x}}{\left(x+1\right)^2}\\ =\dfrac{x+1-2x}{2\sqrt{x}\left(x+1\right)^2}\\ =\dfrac{-x+1}{2\sqrt{x}\left(x+1\right)^2}\)
\(b,y'=\left(\sqrt{x}+1\right)'\left(x^2+2\right)+\left(\sqrt{x}+1\right)\left(x^2+2\right)'\\ =\dfrac{x^2+2}{2\sqrt{x}}+\left(\sqrt{x}+1\right)\cdot2x\)
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: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)
\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} x = 1\)
Vậy \(f'\left( 1 \right) = 1\)
b) \(f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to - 1} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{ - {x^3} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{ - \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - x + 1} \right) = 3\)
Vậy \(f'\left( { - 1} \right) = 3\)
B
1
20 tháng 8 2023
a: \(y'=\left(x^2-x\right)'=2x-1\)
\(y''=\left(2x-1\right)'=2\)
b: \(y'=\left(cosx\right)'=-sinx\)
\(y''=\left(-sinx\right)'=-cosx\)
a: \(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{x^2+1-x_0^2-1}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{\left(x-x0\right)\left(x+x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}x+x0=x0+x0=2x0\)
b: \(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{kx+c-k\cdot x0-c}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{k\left(x-x0\right)}{x-x0}\)
=\(\lim\limits_{x\rightarrow x0}k=k\)