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Câu hỏi: Tính đạo hàm của hàm số $y=\ln \left( {{x}^{4}}+4{{x}^{3}}-3 \right)$.
A. ${y}'=\dfrac{1}{{{x}^{4}}+4{{x}^{3}}-3}$.
B. ${y}'=\dfrac{1}{4{{x}^{3}}+12{{x}^{2}}}$.
C. ${y}'=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{\left( {{x}^{4}}+4{{x}^{3}}-3 \right)}^{2}}}$.
D. ${y}'=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{x}^{4}}+4{{x}^{3}}-3}$.
Ta có $y'=\left[ \ln \left( {{x}^{4}}+4{{x}^{3}}-3 \right) \right]'=\dfrac{\left( {{x}^{4}}+4{{x}^{3}}-3 \right)'}{{{x}^{4}}+4{{x}^{3}}-3}=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{x}^{4}}+4{{x}^{3}}-3}.$
A. ${y}'=\dfrac{1}{{{x}^{4}}+4{{x}^{3}}-3}$.
B. ${y}'=\dfrac{1}{4{{x}^{3}}+12{{x}^{2}}}$.
C. ${y}'=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{\left( {{x}^{4}}+4{{x}^{3}}-3 \right)}^{2}}}$.
D. ${y}'=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{x}^{4}}+4{{x}^{3}}-3}$.
Ta có $y'=\left[ \ln \left( {{x}^{4}}+4{{x}^{3}}-3 \right) \right]'=\dfrac{\left( {{x}^{4}}+4{{x}^{3}}-3 \right)'}{{{x}^{4}}+4{{x}^{3}}-3}=\dfrac{4{{x}^{3}}+12{{x}^{2}}}{{{x}^{4}}+4{{x}^{3}}-3}.$
Đáp án D.