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Câu hỏi: Đạo hàm của hàm số $y=\dfrac{{{3}^{x}}-1}{{{5}^{x}}}$ à
A. ${y}'={{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}+{{\left( \dfrac{1}{5} \right)}^{x}}\ln 5$
B. ${y}'=x{{\left( \dfrac{3}{5} \right)}^{x-1}}-x{{\left( \dfrac{1}{5} \right)}^{x-1}}$
C. ${y}'={{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}-{{\left( \dfrac{1}{5} \right)}^{x}}\ln 5$
D. ${y}'=x{{\left( \dfrac{3}{5} \right)}^{x-1}}+x{{\left( \dfrac{1}{5} \right)}^{x-1}}$
A. ${y}'={{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}+{{\left( \dfrac{1}{5} \right)}^{x}}\ln 5$
B. ${y}'=x{{\left( \dfrac{3}{5} \right)}^{x-1}}-x{{\left( \dfrac{1}{5} \right)}^{x-1}}$
C. ${y}'={{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}-{{\left( \dfrac{1}{5} \right)}^{x}}\ln 5$
D. ${y}'=x{{\left( \dfrac{3}{5} \right)}^{x-1}}+x{{\left( \dfrac{1}{5} \right)}^{x-1}}$
Ta có: $y=\dfrac{{{3}^{x}}-1}{{{5}^{x}}}={{\left( \dfrac{3}{5} \right)}^{x}}-{{\left( \dfrac{1}{5} \right)}^{x}}\Rightarrow {y}'{{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}-{{\left( \dfrac{1}{5} \right)}^{x}}\ln \dfrac{1}{5}={{\left( \dfrac{3}{5} \right)}^{x}}\ln \dfrac{3}{5}+{{\left( \dfrac{1}{5} \right)}^{x}}\ln 5$
Đáp án A.