Đáp án:
\(y' = \dfrac{{8x}}{{\sqrt {8{x^2} + 2} .{{\cos }^2}\left( {\sqrt {8{x^2} + 2} } \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
y = \dfrac{{{x^3}}}{8} - \dfrac{{{x^5}}}{8} + x - 1\\
y' = \dfrac{3}{8}{x^2} - \dfrac{5}{8}{x^4} + 1\\
y = 8\sin x - 3\cos 8x\\
y' = 8.\cos x - 3.8.\left( { - 1} \right)\sin 8x\\
= 8\cos x + 24\sin 8x\\
y = \tan \sqrt {8{x^2} + 2} \\
y' = \left( {16x} \right).\dfrac{1}{{2\sqrt {8{x^2} + 2} }}.\dfrac{1}{{{{\cos }^2}\left( {\sqrt {8{x^2} + 2} } \right)}}\\
= \dfrac{{8x}}{{\sqrt {8{x^2} + 2} .{{\cos }^2}\left( {\sqrt {8{x^2} + 2} } \right)}}
\end{array}\)
