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Câu hỏi: Tính đạo hàm của hàm số $y={{\left( \dfrac{1}{2022} \right)}^{x}}$.
A. ${y}'={{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
B. ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
C. ${y}'=x{{\left( \dfrac{1}{2022} \right)}^{x-1}}\ln 2022$.
D. ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\dfrac{1}{\ln 2022}$.
A. ${y}'={{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
B. ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
C. ${y}'=x{{\left( \dfrac{1}{2022} \right)}^{x-1}}\ln 2022$.
D. ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\dfrac{1}{\ln 2022}$.
Ta có: ${{\left( {{a}^{x}} \right)}^{\prime }}={{a}^{x}}\ln a$, với $0<a\ne 1$.
Do đó ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
Do đó ${y}'=-{{\left( \dfrac{1}{2022} \right)}^{x}}\ln 2022$.
Đáp án B.