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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/ \(y'=\frac{4\sqrt{x^2+2}-\left(4x+1\right)\left(\sqrt{x^2+2}\right)'}{x^2+2}=\frac{4\sqrt{x^2+2}-\left(4x+1\right).\frac{x}{\sqrt{x^2+2}}}{x^2+2}\)
\(=\frac{4\left(x^2+2\right)-\left(4x^2+x\right)}{\left(x^2+2\right)\sqrt{x^2+2}}=\frac{8-x}{\left(x^2+2\right)\sqrt{x^2+2}}\)
Ta có : \(y'=x+\frac{1}{2}\left(\sqrt{x^2+1}+x\frac{x}{\sqrt{x^2+1}}\right)+\frac{\frac{1+\frac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}}{\sqrt{x+\sqrt{x^2+1}}}\)
\(=x+\frac{2x^2+1}{2\sqrt{x^2+1}}+\frac{x+\sqrt{x^2+1}}{2\left(x+\sqrt{x^2+1}\right)\sqrt{x^2+1}}=x+\frac{2x^2+1}{2\sqrt{x^2+1}}+\frac{1}{2\sqrt{x^2+1}}\)
\(=x+\frac{2\left(x^2+1\right)}{2\sqrt{x^2+1}}=x+\sqrt{x^2+1}\)
\(\Rightarrow\begin{cases}xy'+\ln y'=x\left(x+\sqrt{x^2+1}\right)+\ln\left(x+\sqrt{x^2+1}\right)=x^2+x\sqrt{x^2+1}+\ln\left(x+\sqrt{x^2+1}\right)\\2y=x^2+x\sqrt{x^2+1}+2\ln\sqrt{x+\sqrt{x^2+1}}=x^2+x\sqrt{x^2+1}+\ln\left(x+\sqrt{x^2+1}\right)\end{cases}\)
\(\Rightarrow2y=xy'+\ln y'\)\(\Rightarrow\) Điều phải chứng minh
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
.
a/ \(y'=42\left(2x+3\right)^{20}\left(x-4\right)^{23}+23\left(x-4\right)^{22}\left(2x+3\right)^{21}\)
b/ \(y=\frac{1}{x\sqrt{x}}=\frac{1}{\sqrt{x^3}}=x^{-\frac{3}{2}}\Rightarrow y'=-\frac{3}{2}x^{-\frac{5}{2}}=-\frac{3}{2x^2\sqrt{x}}\)
c/ \(y'=\frac{\left(x+\frac{1}{x}\right)'}{2\sqrt{\frac{x^2+1}{x}}}=\frac{1-\frac{1}{x^2}}{2\sqrt{\frac{x^2+1}{x}}}=\frac{\left(x^2-1\right)\sqrt{x}}{2x^2\sqrt{x^2+1}}\)
d/ \(y=x^2+x^{\frac{3}{2}}+1\Rightarrow y'=2x+\frac{3}{2}x^{\frac{1}{2}}=2x+\frac{3}{2}\sqrt{x}\)
e/ \(y'=\frac{\sqrt{1-x}+\frac{1+x}{2\sqrt{1-x}}}{1-x}=\frac{3-x}{2\left(1-x\right)\sqrt{1-x}}\)
f/ \(y'=\frac{\sqrt{a^2-x^2}+\frac{x^2}{\sqrt{a^2-x^2}}}{a^2-x^2}=\frac{a^2}{a^2-x^2}\)
\(y'=\dfrac{\left(x+\sqrt{1+x^2}\right)'}{2\sqrt{x+\sqrt{1+x^2}}}=\dfrac{1+\dfrac{x}{\sqrt{1+x^2}}}{2\sqrt{x+\sqrt{1+x^2}}}\)
\(\Rightarrow2\sqrt{1+x^2}.y'=\dfrac{2\sqrt{1+x^2}\left(1+\dfrac{x}{\sqrt{1+x^2}}\right)}{2\sqrt{x+\sqrt{1+x^2}}}\)
\(=\dfrac{\sqrt{1+x^2}+x}{\sqrt{x+\sqrt{1+x^2}}}=\sqrt{x+\sqrt{1+x^2}}=y\) (đpcm)