Đáp án:a/
$(2x - 1).{e^{{x^2} - x}}$
b/
$-9\ln 3.{\left( {\frac{1}{{27}}} \right)^{\sin 3x}}.\cos 3x$
Giải thích các bước giải:a/
$y' = ({e^{{x^2} - x}})' = ({e^{{x^2} - x}})({x^2} - x)' = (2x - 1).{e^{{x^2} - x}}$
b/
$\begin{array}{l}
y = {3^{\cos '3x}} = {3^{ - 3\sin 3x}} = {\left( {\frac{1}{{27}}} \right)^{\sin 3x}}\\
= > y' = {\left( {\frac{1}{{27}}} \right)^{\sin 3x}}.\ln \left( {\frac{1}{{27}}} \right).\sin '3x\\
= {\left( {\frac{1}{{27}}} \right)^{\sin 3x}}.( - 3)\ln 3.3\cos (3x)\\
= - 9\ln 3.{\left( {\frac{1}{{27}}} \right)^{\sin 3x}}.\cos 3x
\end{array}$
