Giải bài 2 trang 61 – SGK môn Giải tích lớp 12
Tính đạo hàm của các hàm số:
a) \(y={{\left( 2{{x}^{2}}-x+1 \right)}^{\frac{1}{3}}};\)
b) \(y={{\left( 4-x-{{x}^{2}} \right)}^{\frac{1}{4}}};\)
c) \(y={{\left( 3x+1 \right)}^{\frac{\pi }{2}}};\)
d) \(y={{\left( 5-x \right)}^{\sqrt{3}}}\)
a) \(y={{\left( 2{{x}^{2}}-x+1 \right)}^{\frac{1}{3}}};\)
\(y'=\dfrac{1}{3}\left( 2{{x}^{2}}-x+1 \right)'{{\left( 2{{x}^{2}}-x+1 \right)}^{\frac{1}{3}-1}}=\dfrac{1}{3}\left( 4x-1 \right){{\left( 2{{x}^{2}}-x+1 \right)}^{\frac{-2}{3}}}\)
b) \(y={{\left( 4-x-{{x}^{2}} \right)}^{\frac{1}{4}}};\)
\(y'=\dfrac{1}{4}\left( 4-x-{{x}^{2}} \right)'{{\left( 4-x-{{x}^{2}} \right)}^{\frac{1}{4}-1}}=-\dfrac{1}{4}\left( 1+2x \right){{\left( 4-x-{{x}^{2}} \right)}^{\frac{-3}{4}}}\)
c) \(y={{\left( 3x+1 \right)}^{\frac{\pi }{2}}};\)
\(y'=\dfrac{\pi }{2}\left( 3x+1 \right)'{{\left( 3x+1 \right)}^{\frac{\pi }{2}-1}}=\dfrac{3\pi }{2}{{\left( 3x+1 \right)}^{\frac{\pi }{2}-1}}\)
d) \(y={{\left( 5-x \right)}^{\sqrt{3}}}\)
\(y'=\sqrt{3}\left( 5-x \right)'{{\left( 5-x \right)}^{\sqrt{3}-1}}=-\sqrt{3}{{\left( 5-x \right)}^{\sqrt{3}-1}}\)
Ghi nhớ: Công thức tính đạo hàm hàm hợp: \(\left( {{u}^{\alpha }} \right)'=\alpha .{{u}^{\alpha -1}}.u'\).