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16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
a) =
=
.
b) =
=
.
c) =
=
.
d) y' =\(\dfrac{\left(x^2+7x+3\right)'\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(x^2-3x\right)'}{\left(x^2-3x\right)^2}\)=\(\dfrac{\left(2x+7\right)\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(2x-3\right)}{\left(x^2-3x\right)^2}\)=\(\dfrac{-2x^2-6x+9}{\left(x^2-3x\right)^2}\)
a/ \(y'=\frac{\left(2x^2-5x+2\right)'}{2\sqrt{2x^2-5x+2}}=\frac{4x-5}{2\sqrt{2x^2-5x+2}}\)
b/ \(y'=\frac{\left(x+\sqrt{x}\right)'}{2\sqrt{x+\sqrt{x}}}=\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}=\frac{2\sqrt{x}+1}{4\sqrt{x^2+x\sqrt{x}}}\)
c/ \(y'=\sqrt{x^2+3}+\left(x-2\right).\frac{\left(x^2+3\right)'}{2\sqrt{x^2+3}}=\frac{2x^2-2x+3}{\sqrt{x^2+3}}\)
d/ \(y'=3\left(1+\sqrt{1-2x}\right)^2.\left(1+\sqrt{1-2x}\right)'=\frac{-3\left(1+\sqrt{1-2x}\right)^2}{\sqrt{1-2x}}\)
e/ \(y'=\frac{1}{2}\sqrt{\frac{x-1}{x^3}}\left(\frac{x^3}{x-1}\right)'=\frac{1}{2}\sqrt{\frac{x-1}{x^3}}\left(\frac{x^2\left(x-1\right)-x^3}{\left(x-1\right)^2}\right)=\frac{-x^2}{2\left(x-1\right)^2}\sqrt{\frac{x-1}{x^3}}\)
f/ \(y'=\frac{4\sqrt{x^2+2}-\left(4x+1\right)\left(\sqrt{x^2+2}\right)'}{x^2+2}=\frac{4\sqrt{x^2+2}-\left(4x+1\right).\frac{x}{\sqrt{x^2+2}}}{x^2+2}\)
\(=\frac{4\left(x^2+2\right)-\left(4x^2+x\right)}{\left(x^2+2\right)\sqrt{x^2+2}}=\frac{8-x}{\left(x^2+2\right)\sqrt{x^2+2}}\)
a) Cách 1: y' = (9 -2x)'(2x3- 9x2 +1) +(9 -2x)(2x3- 9x2 +1)' = -2(2x3- 9x2 +1) +(9 -2x)(6x2 -18x) = -16x3 +108x2 -162x -2.
Cách 2: y = -4x4 +36x3 -81x2 -2x +9, do đó
y' = -16x3 +108x2 -162x -2.
b) y' = .(7x -3) +
(7x -3)'=
(7x -3) +7
.
c) y' = (x -2)'√(x2 +1) + (x -2)(√x2 +1)' = √(x2 +1) + (x -2) = √(x2 +1) + (x -2)
= √(x2 +1) +
=
.
d) y' = 2tanx.(tanx)' - (x2)' =
.
e) y' = sin
=
sin
.
