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\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
Ta có
f ( x ) = ( x + 2 ) ( x − 3 ) = x 2 − x − 6 ⇒ f ' x = 2 x − 1
Chọn đáp án C
1.
\(f'\left(x\right)=3x^2-6mx+3\left(2m-1\right)\)
\(f'\left(x\right)-6x=3x^2-3.2\left(m+1\right)x+3\left(2m-1\right)>0\)
\(\Leftrightarrow x^2-2\left(m+1\right)x+2m-1>0\)
\(\Leftrightarrow x^2-2x-1>2m\left(x-1\right)\)
Do \(x>2\Rightarrow x-1>0\) nên BPT tương đương:
\(\dfrac{x^2-2x-1}{x-1}>2m\Leftrightarrow\dfrac{\left(x-1\right)^2-2}{x-1}>2m\)
Đặt \(t=x-1>1\Rightarrow\dfrac{t^2-2}{t}>2m\Leftrightarrow f\left(t\right)=t-\dfrac{2}{t}>2m\)
Xét hàm \(f\left(t\right)\) với \(t>1\) : \(f'\left(t\right)=1+\dfrac{2}{t^2}>0\) ; \(\forall t\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow f\left(t\right)>f\left(1\right)=-1\Rightarrow\) BPT đúng với mọi \(t>1\) khi \(2m< -1\Rightarrow m< -\dfrac{1}{2}\)
2.
Thay \(x=0\) vào giả thiết:
\(f^3\left(2\right)-2f^2\left(2\right)=0\Leftrightarrow f^2\left(2\right)\left[f\left(2\right)-2\right]=0\Rightarrow\left[{}\begin{matrix}f\left(2\right)=0\\f\left(2\right)=2\end{matrix}\right.\)
Đạo hàm 2 vế giả thiết:
\(-3f^2\left(2-x\right).f'\left(2-x\right)-12f\left(2+3x\right).f'\left(2+3x\right)+2x.g\left(x\right)+x^2.g'\left(x\right)+36=0\) (1)
Thế \(x=0\) vào (1) ta được:
\(-3f^2\left(2\right).f'\left(2\right)-12f\left(2\right).f'\left(2\right)+36=0\)
\(\Leftrightarrow f^2\left(2\right).f'\left(2\right)+4f\left(2\right).f'\left(2\right)-12=0\) (2)
Với \(f\left(2\right)=0\) thế vào (2) \(\Rightarrow-12=0\) ko thỏa mãn (loại)
\(\Rightarrow f\left(2\right)=2\)
Thế vào (2):
\(4f'\left(2\right)+8f'\left(2\right)-12=0\Leftrightarrow f'\left(2\right)=1\)
\(\Rightarrow A=3.2+4.1\)
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. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
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. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)
2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)
3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)
1/ \(y'=\dfrac{\left(\sqrt{x+1}\right)'x-x'\sqrt{x+1}}{x^2}=\dfrac{\dfrac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}=\dfrac{-x-2}{2x^2\sqrt{x+1}}\)
2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)
3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)
4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)
\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)
5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học
1) \(f\left(x\right)=2x-5\)
\(f'\left(x\right)=2\)
\(\Rightarrow f'\left(4\right)=2\)
2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)
\(\Rightarrow y'=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)
3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)
\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)
Áp dụng công thức 1 v ' = − v ' v 2 .
Ta có: . f ' ( x ) = − x 2 − 1 ' x 2 − 1 2 = − 2 x x 2 − 1 2
Chọn đáp án B