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Ta có : \(y'=\frac{\left(2x^3+1\right)'}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)
Ta có :
\(y'=\frac{\left(2x^3+1\right)'}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)
Ta có \(y'=e^{\sqrt[3]{x^2+1}-x}\left(\sqrt[3]{x^2+1}-x\right)+3^{3x-1}\left(3x-1\right)'\ln3\)
\(=e^{\sqrt[3]{x^2+1}-x}\left(\frac{2x}{3\sqrt[3]{\left(x^2+1\right)^2}}-1\right)+3^{3x}\ln3\)
\(y=\sqrt{x\sqrt[3]{x\sqrt[4]{x}}}=x^{\frac{1}{2}}.x^{\frac{1}{2}.\frac{1}{3}}.x^{\frac{1}{2}.\frac{1}{3}.\frac{1}{4}}=x^{\frac{17}{24}}\)
\(\Rightarrow y'=\frac{17}{24}.x^{\frac{17}{24}-1}=\frac{17}{24}.x^{\frac{-7}{24}}=\frac{17}{24\sqrt[24]{x^7}}\)
Ta có \(y=\log_3\left(\frac{x^2-2x+3}{x^2+2x+3}\right)=\log_3\left(x^2-2x+3\right)-\log_3\left(x^2+2x+3\right)\)
\(\Rightarrow y'=\frac{2x-2}{\left(x^2-2x+3\right)\ln3}-\frac{2x-2}{\left(x^2+2x+3\right)\ln3}=\frac{4x^2-12}{\left(x^4+2x^2+9\right)\ln3}\)
a) \(f\left(x\right)=\sin^3x.\sin3x=\sin3x\left(\frac{3\sin x-\sin3x}{4}\right)=\frac{3}{4}\sin3x.\sin x-\frac{1}{4}\sin^23x\)
= \(\frac{3}{8}\left(\cos2x-\cos4x\right)-\frac{1}{8}\left(1-\cos6x\right)=\frac{3}{8}\cos2x+\frac{1}{8}\cos6x-\frac{3}{8}\cos4x-\frac{1}{8}\)
Do đó :
\(I=\int f\left(x\right)dx=\int\left(\frac{3}{8}\cos2x+\frac{1}{8}\cos6x-\frac{3}{8}\cos4x-\frac{1}{8}\right)dx=\frac{3}{16}\sin2x+\frac{1}{48}\sin6x-\frac{3}{32}\sin4x-\frac{1}{8}x+C\)
b) Ta biến đổi :
\(f\left(x\right)=\sin^3x.\cos3x+\cos^3x.\sin3x=\cos3x\left(\frac{3\sin x-\sin3x}{4}\right)+\sin3x\left(\frac{\cos3x+3\cos x}{4}\right)\)
\(=\frac{3}{4}\left(\cos3x\sin x+\sin3x\cos x\right)=\frac{3}{4}\sin4x\)
Do đó : \(I=\int f\left(x\right)dx=\frac{3}{4}\int\sin4xdx=-\frac{3}{16}\cos4x+C\)
\(y=\left(\frac{2}{3}\right)^x-\left(\frac{1}{6}\right)^x\Rightarrow y'=\frac{1}{x\ln\frac{2}{3}}+\frac{1}{x\ln6}\)
\(=\frac{2\ln2}{x\ln6\left(\ln2-\ln3\right)}\)
\(y'=\frac{3\cos3x}{10\sqrt[10]{\sin^93x}}\)