Giải bài 58 trang 117 SGK giải tích nâng cao 12

a) Áp dụng công thức \((u^n)'=nu^{n-1}.u'\) ta có

\(y={{\left( 2x+1 \right)}^{\pi }} \\ y'=2\pi {{\left( 2x+1 \right)}^{\pi -1}} \)

b) Áp dụng công thức \(\left( \sqrt[n]{u} \right)'=\dfrac{u'}{n\sqrt[n]{{{u}^{n-1}}}}\) ta có

\(y=\sqrt[5]{{{\ln }^{3}}5x} \\ y'=\dfrac{\left( {{\ln }^{3}}5x \right)'}{5\sqrt[5]{{{\left( {{\ln }^{3}}5x \right)}^{4}}}}=\dfrac{3{{\ln }^{2}}5x.\left( \ln 5x \right)'}{5\sqrt[5]{{{\ln }^{12}}5x}}=\dfrac{3}{5x\sqrt[5]{{{\ln }^{2}}5x}} \\ \)

c) \( y=\sqrt[3]{\dfrac{1+{{x}^{3}}}{1-{{x}^{3}}}}\)

Đặt \(\dfrac{1+{{x}^{3}}}{1-{{x}^{3}}}=u\Rightarrow y'=\dfrac{u'}{3\sqrt[3]{{{u}^{2}}}} \)

\(u'=\dfrac{3{{x}^{2}}\left( 1-{{x}^{3}} \right)-3{{x}^{2}}\left( 1+{{x}^{3}} \right)}{{{\left( 1-{{x}^{3}} \right)}^{2}}}=\dfrac{6{{x}^{2}}}{{{\left( 1-{{x}^{3}} \right)}^{2}}} \\ \Rightarrow y'=\dfrac{2{{x}^{2}}}{{{\left( 1-{{x}^{3}} \right)}^{2}}}.\dfrac{1}{\sqrt[3]{{{\left( \dfrac{1+{{x}^{3}}}{1-{{x}^{3}}} \right)}^{2}}}}=\dfrac{2{{x}^{2}}}{\sqrt[3]{{{\left( 1-{{x}^{3}} \right)}^{4}}{{\left( 1+{{x}^{3}} \right)}^{2}}}} \)

d) \(y={{\left( \dfrac{x}{b} \right)}^{a}}{{\left( \dfrac{a}{x} \right)}^{b}}\)

\(y'={{\left[ {{\left( \dfrac{x}{b} \right)}^{a}} \right]}^{'}}{{\left( \dfrac{a}{x} \right)}^{b}}+{{\left( \dfrac{x}{b} \right)}^{a}}{{\left[ {{\left( \dfrac{a}{x} \right)}^{b}} \right]}^{'}} \\ =\dfrac{a}{b}{{\left( \dfrac{x}{b} \right)}^{a-1}}{{\left( \dfrac{a}{x} \right)}^{b}}+b{{\left( \dfrac{x}{b} \right)}^{a}}{{\left( \dfrac{a}{x} \right)}^{b-1}}\left( -\dfrac{a}{{{x}^{2}}} \right) \\ ={{\left( \dfrac{x}{b} \right)}^{a}}{{\left( \dfrac{a}{x} \right)}^{b}}\dfrac{a-b}{x} \)