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Câu hỏi: Tìm đạo hàm của hàm số: $y={{({{x}^{2}}+1)}^{\dfrac{3}{2}}}$
A. $\dfrac{3}{2}{{(2x)}^{\dfrac{1}{2}}}$
B. $\dfrac{3}{4}{{x}^{-\dfrac{1}{4}}}$
C. $3x{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
D. $\dfrac{3}{2}{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
A. $\dfrac{3}{2}{{(2x)}^{\dfrac{1}{2}}}$
B. $\dfrac{3}{4}{{x}^{-\dfrac{1}{4}}}$
C. $3x{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
D. $\dfrac{3}{2}{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
Áp dụng công thức đạo hàm hợp hàm số lũy thừa : ${{\left( {{\left[ u(x) \right]}^{\alpha }} \right)}^{'}}=\alpha .{{u}^{\alpha -1}}.{{\left[ u(x) \right]}^{'}}$
Ta có : $y'={{\left( {{({{x}^{2}}+1)}^{\dfrac{3}{2}}} \right)}^{'}}=\dfrac{3}{2}.2x.{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}=3x.{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
Ta có : $y'={{\left( {{({{x}^{2}}+1)}^{\dfrac{3}{2}}} \right)}^{'}}=\dfrac{3}{2}.2x.{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}=3x.{{({{x}^{2}}+1)}^{\dfrac{1}{2}}}$
Đáp án C.
