Đạo hàm cấp n của hàm số \(\dfrac{1}{ax+b},\,a\ne 0\) là :
A.\({{y}^{\left( n \right)}}=\dfrac{{{2}^{n}}.{{a}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
B.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}{{a}^{n}}n!}{{{\left( x+1 \right)}^{n+1}}}\)
C.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
D.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}.{{a}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
A.\({{y}^{\left( n \right)}}=\dfrac{{{2}^{n}}.{{a}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
B.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}{{a}^{n}}n!}{{{\left( x+1 \right)}^{n+1}}}\)
C.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
D.\({{y}^{\left( n \right)}}=\dfrac{{{\left( -1 \right)}^{n}}.{{a}^{n}}.n!}{{{\left( ax+b \right)}^{n+1}}}\)
Loga
Hóa Học lớp 12
