1: \(y=2x+cosx\) => \(y'=2-sinx\) => \(y''=2'-\left(sinx\right)'=-cosx\) 2: \(y=sin^3x\) => \(y'=3\cdot sin^2x\cdot\left(sinx\right)'=3\cdot sin^2x\cdot cosx\) => \(y''=3\cdot\left(sin^2x\cdot cosx\right)'\) => \(y''=3\left[\left(sin^2x\right)'\cdot cosx+\left(sin^2x\right)\cdot\left(cosx\right)'\right]\) => \(y''=3\left[2\cdot sinx\cdot\left(sinx\right)'\cdot cosx+sin^2x\cdot\left(-sinx\right)\right]\) => \(y''=3\left[2\cdot sinx\cdot cosx\cdot sinx-sin^3x\right]\) => \(y''=6\cdot sin^2x\cdot cosx-3\cdot sin^3x\) 3: \(y=2\cdot sin2x-cos\left(x+\dfrac{\Omega}{3}\right)\) => \(y'=2\cdot\left(2x\right)'\cdot\left(cos2x\right)-\left(-1\right)\cdot\left(x+\dfrac{\Omega}{3}\right)'\cdot sin\left(x+\dfrac{\Omega}{3}\right)\) => \(y'=4\cdot cos2x+sin\left(x+\dfrac{\Omega}{3}\right)\) => \(y''=4\cdot\left(-1\right)\cdot\left(2x\right)'\cdot sin2x+\left(x+\dfrac{\Omega}{3}\right)'\cdot cos\left(x+\dfrac{\Omega}{3}\right)\) => \(y''=-8\cdot sin2x+cos\left(x+\dfrac{\Omega}{3}\right)\) 4: \(y=\sqrt{x^2+1}\) => \(y'=\dfrac{\left(x^2+1\right)'}{2\sqrt{x^2+1}}=\dfrac{2x}{2\sqrt{x^2+1}}=\dfrac{x}{\sqrt{x^2+1}}\) => \(y''=\dfrac{x'\cdot\sqrt{x^2+1}-x\cdot\left(\sqrt{x^2+1}\right)'}{x^2+1}\) => \(y''=\dfrac{\sqrt{x^2+1}-x\cdot\dfrac{x}{\sqrt{x^2+1}}}{x^2+1}\) => \(y''=\dfrac{x^2+1-x^2}{\sqrt{x^2+1}\cdot\left(x^2+1\right)}=\dfrac{1}{\left(x^2+1\right)\cdot\sqrt{x^2+1}}\) 5: \(y=x\cdot cosx\) => \(y'=x'\cdot cosx+x\cdot\left(cosx\right)'=cosx-sinx\cdot x\) => \(y''=\left(cosx\right)'-\left(sinx\cdot x\right)'\) => \(y''=-sinx-\left[\left(sinx\right)'\cdot x+sinx\cdot x'\right]\) => \(y''=-sinx-cosx\cdot x-sinx\) => \(y''=-2\cdot sinx-cosx\cdot x\) 6: \(y=\dfrac{x+2}{x-3}\) => \(y'=\dfrac{\left(x+2\right)'\left(x-3\right)-\left(x+2\right)\left(x-3\right)'}{\left(x-3\right)^2}\) => \(y''=\dfrac{x-3-x-2}{\left(x-3\right)^2}=\dfrac{-5}{\left(x-3\right)^2}\) => \(y''=\dfrac{\left(-5\right)'\cdot\left(x-3\right)^2-\left(-5\right)\cdot\left[\left(x-3\right)^2\right]'}{\left(x-3\right)^4}\) => \(y''=\dfrac{5\cdot\left(x^2-6x+9\right)'}{\left(x-3\right)^4}\) => \(y''=\dfrac{5\left(2x-6\right)}{\left(x-3\right)^4}=\dfrac{10}{\left(x-3\right)^3}\) Câu trả lời: 1: \(y=2x+cosx\) => \(y'=2-sinx\) => \(y''=2'-\left(sinx\right)'=-cosx\) 2: \(y=sin^3x\) => \(y'=3\cdot sin^2x\cdot\left(sinx\right)'=3\cdot sin^2x\cdot cosx\) => \(y''=3\cdot\left(sin^2x\cdot cosx\right)'\) => \(y''=3\left[\left(sin^2x\right)'\cdot cosx+\left(sin^2x\right)\cdot\left(cosx\right)'\right]\) => \(y''=3\left[2\cdot sinx\cdot\left(sinx\right)'\cdot cosx+sin^2x\cdot\left(-sinx\right)\right]\) => \(y''=3\left[2\cdot sinx\cdot cosx\cdot sinx-sin^3x\right]\) => \(y''=6\cdot sin^2x\cdot cosx-3\cdot sin^3x\) 3: \(y=2\cdot sin2x-cos\left(x+\dfrac{\Omega}{3}\right)\) => \(y'=2\cdot\left(2x\right)'\cdot\left(cos2x\right)-\left(-1\right)\cdot\left(x+\dfrac{\Omega}{3}\right)'\cdot sin\left(x+\dfrac{\Omega}{3}\right)\) => \(y'=4\cdot cos2x+sin\left(x+\dfrac{\Omega}{3}\right)\) => \(y''=4\cdot\left(-1\right)\cdot\left(2x\right)'\cdot sin2x+\left(x+\dfrac{\Omega}{3}\right)'\cdot cos\left(x+\dfrac{\Omega}{3}\right)\) => \(y''=-8\cdot sin2x+cos\left(x+\dfrac{\Omega}{3}\right)\) 4: \(y=\sqrt{x^2+1}\) => \(y'=\dfrac{\left(x^2+1\right)'}{2\sqrt{x^2+1}}=\dfrac{2x}{2\sqrt{x^2+1}}=\dfrac{x}{\sqrt{x^2+1}}\) => \(y''=\dfrac{x'\cdot\sqrt{x^2+1}-x\cdot\left(\sqrt{x^2+1}\right)'}{x^2+1}\) => \(y''=\dfrac{\sqrt{x^2+1}-x\cdot\dfrac{x}{\sqrt{x^2+1}}}{x^2+1}\) => \(y''=\dfrac{x^2+1-x^2}{\sqrt{x^2+1}\cdot\left(x^2+1\right)}=\dfrac{1}{\left(x^2+1\right)\cdot\sqrt{x^2+1}}\) 5: \(y=x\cdot cosx\) => \(y'=x'\cdot cosx+x\cdot\left(cosx\right)'=cosx-sinx\cdot x\) => \(y''=\left(cosx\right)'-\left(sinx\cdot x\right)'\) => \(y''=-sinx-\left[\left(sinx\right)'\cdot x+sinx\cdot x'\right]\) => \(y''=-sinx-cosx\cdot x-sinx\) => \(y''=-2\cdot sinx-cosx\cdot x\) 6: \(y=\dfrac{x+2}{x-3}\) => \(y'=\dfrac{\left(x+2\right)'\left(x-3\right)-\left(x+2\right)\left(x-3\right)'}{\left(x-3\right)^2}\) => \(y''=\dfrac{x-3-x-2}{\left(x-3\right)^2}=\dfrac{-5}{\left(x-3\right)^2}\) => \(y''=\dfrac{\left(-5\right)'\cdot\left(x-3\right)^2-\left(-5\right)\cdot\left[\left(x-3\right)^2\right]'}{\left(x-3\right)^4}\) => \(y''=\dfrac{5\cdot\left(x^2-6x+9\right)'}{\left(x-3\right)^4}\) => \(y''=\dfrac{5\left(2x-6\right)}{\left(x-3\right)^4}=\dfrac{10}{\left(x-3\right)^3}\)

Hà anh

9 tháng 5 2024 - olm

giúp em với

#Hỏi cộng đồng OLM #Toán lớp 11

Nguyễn Lê Phước Thịnh

9 tháng 5 2024

1: \(y=2x+cosx\)

=>\(y'=2-sinx\)

=>\(y''=2'-\left(sinx\right)'=-cosx\)

2: \(y=sin^3x\)

=>\(y'=3\cdot sin^2x\cdot\left(sinx\right)'=3\cdot sin^2x\cdot cosx\)

=>\(y''=3\cdot\left(sin^2x\cdot cosx\right)'\)

=>\(y''=3\left[\left(sin^2x\right)'\cdot cosx+\left(sin^2x\right)\cdot\left(cosx\right)'\right]\)

=>\(y''=3\left[2\cdot sinx\cdot\left(sinx\right)'\cdot cosx+sin^2x\cdot\left(-sinx\right)\right]\)

=>\(y''=3\left[2\cdot sinx\cdot cosx\cdot sinx-sin^3x\right]\)

=>\(y''=6\cdot sin^2x\cdot cosx-3\cdot sin^3x\)

3: \(y=2\cdot sin2x-cos\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y'=2\cdot\left(2x\right)'\cdot\left(cos2x\right)-\left(-1\right)\cdot\left(x+\dfrac{\Omega}{3}\right)'\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y'=4\cdot cos2x+sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y''=4\cdot\left(-1\right)\cdot\left(2x\right)'\cdot sin2x+\left(x+\dfrac{\Omega}{3}\right)'\cdot cos\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y''=-8\cdot sin2x+cos\left(x+\dfrac{\Omega}{3}\right)\)

4: \(y=\sqrt{x^2+1}\)

=>\(y'=\dfrac{\left(x^2+1\right)'}{2\sqrt{x^2+1}}=\dfrac{2x}{2\sqrt{x^2+1}}=\dfrac{x}{\sqrt{x^2+1}}\)

=>\(y''=\dfrac{x'\cdot\sqrt{x^2+1}-x\cdot\left(\sqrt{x^2+1}\right)'}{x^2+1}\)

=>\(y''=\dfrac{\sqrt{x^2+1}-x\cdot\dfrac{x}{\sqrt{x^2+1}}}{x^2+1}\)

=>\(y''=\dfrac{x^2+1-x^2}{\sqrt{x^2+1}\cdot\left(x^2+1\right)}=\dfrac{1}{\left(x^2+1\right)\cdot\sqrt{x^2+1}}\)

5: \(y=x\cdot cosx\)

=>\(y'=x'\cdot cosx+x\cdot\left(cosx\right)'=cosx-sinx\cdot x\)

=>\(y''=\left(cosx\right)'-\left(sinx\cdot x\right)'\)

=>\(y''=-sinx-\left[\left(sinx\right)'\cdot x+sinx\cdot x'\right]\)

=>\(y''=-sinx-cosx\cdot x-sinx\)

=>\(y''=-2\cdot sinx-cosx\cdot x\)

6: \(y=\dfrac{x+2}{x-3}\)

=>\(y'=\dfrac{\left(x+2\right)'\left(x-3\right)-\left(x+2\right)\left(x-3\right)'}{\left(x-3\right)^2}\)

=>\(y''=\dfrac{x-3-x-2}{\left(x-3\right)^2}=\dfrac{-5}{\left(x-3\right)^2}\)

=>\(y''=\dfrac{\left(-5\right)'\cdot\left(x-3\right)^2-\left(-5\right)\cdot\left[\left(x-3\right)^2\right]'}{\left(x-3\right)^4}\)

=>\(y''=\dfrac{5\cdot\left(x^2-6x+9\right)'}{\left(x-3\right)^4}\)

=>\(y''=\dfrac{5\left(2x-6\right)}{\left(x-3\right)^4}=\dfrac{10}{\left(x-3\right)^3}\)

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x	– 100 000	– 10 000	– 1 000	– 100	– 10
y = f(x)	?	?	?	–0,01	–0,1

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