Đạo hàm của hàm số y = 2 − 3 x 2 x + 1 2 bằng biểu thức nào sau đây?
A. − 14 2 x + 1 2 . 2 − 3 x 2 x + 1
B. − 4 2 x + 1 2 . 2 − 3 x 2 x + 1
C. 16 2 x + 1 2 . 2 − 3 x 2 x + 1
D. 2 2 − 3 x 2 x + 1
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Ta có
f ( x ) = ( x + 2 ) ( x − 3 ) = x 2 − x − 6 ⇒ f ' x = 2 x − 1
Chọn đáp án C
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
Áp dụng công thức u v ' = u ' . v − v ' . u v 2 .
Ta có:
y ' = x 2 + x + 3 ' x 2 + x − 1 − x 2 + x − 1 ' x 2 + x + 3 x 2 + x − 1 2
= ( 2 x + 1 ) ( x 2 + x − 1 ) − ( 2 x + 1 ) . ( x 2 + x + 3 ) ( x 2 + x − 1 ) 2 = ( 2 x + 1 ) . ( x 2 + x − 1 − x 2 − x − 3 ) ( x 2 + x − 1 ) 2 = − 4 2 x + 1 x 2 + x − 1 2
Chọn đáp án B
Chọn C
y' = (2x +1)(3x - 2)2 + x.2.(3x - 2)2 + x(2x +1).2.(3x - 2).3
(3x - 2)[(2x +1)(3x - 2) + 2x.(3x - 2) + 6x(2x + 1)]
= (3x - 2)(24x2 + x - 2)
Áp dụng công thức u n ' = n u n − 1 . u '
và a x + b c x + d ' = a d − b c c x + d 2 .
Ta có:
y ' = 2 2 − 3 x 2 x + 1 . 2 − 3 x 2 x + 1 ' = 2 2 − 3 x 2 x + 1 . − 3. ( 2 x + 1 ) − 2. ( 2 − 3 x ) ( 2 x + 1 ) 2
= 2 2 − 3 x 2 x + 1 . − 6 x − 3 − 4 + 6 x 2 x + 1 2 = 2 − 3 x 2 x + 1 . − 14 2 x + 1 2
Chọn đáp án A