Đạo hàm của hàm số: f x = x 2 + 1 a 2 + x 2 (a là hằng số) bằng biểu thức nào sau đây?
A. 2 x a 2 - x 2
B. x ( 3 x 2 + 2 a 2 + 1 ) ( a 2 + x 2 ) . a 2 + x 2
C. 2 x 2 a - 2 x
D. x 2 a 2 - x 2 + 1 a 2 - x 2 a 2 - x 2
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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
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
Ta có
f ' x = 2. ( 2 x − 1 ) − 2. ( 2 x − 3 ) ( 2 x − 1 ) 2 = 4 2 x − 1 2
Chọn đáp án D
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
Chọn B
y ' = 2 x . a + 2 x 2 + ( x 2 + 1 ) . 1 2 a 2 + x 2 . ( a 2 + x 2 ) ' a 2 + x 2 = 2 x . a + 2 x 2 + x ( x 2 + 1 ) a 2 + x 2 a 2 + x 2 = 2 x ( a 2 + x 2 ) + x ( x 2 + 1 ) ( a 2 + x 2 ) . a 2 + x 2 = x ( 3 x 2 + 2 a 2 + 1 ) ( a 2 + x 2 ) . a 2 + x 2