Đáp án: a, $=8\frac{\sin ^{3}(\frac{\pi }{6}+2x)}{\cos ^{5}(\frac{\pi }{6}+2x)}$
b, $=20\frac{\cos ^{3}(5x-1)}{\sin ^{5}(5x-1)}$
Giải thích các bước giải:
a)$ y'=4\tan ^{3}(\frac{\pi }{6}+2x)\tan '(\frac{\pi }{6}+2x)$
$=4\tan ^{3}(\frac{\pi }{6}+2x)(\frac{\pi }{6}+2x)'\frac{1}{\cos ^{2}(\frac{\pi }{6}+2x)}$
$=8\frac{\sin ^{3}(\frac{\pi }{6}+2x)}{\cos ^{5}(\frac{\pi }{6}+2x)}$
b)
$y'=4\cot ^{3}(5x-1)\cot '(5x-1)$
$=4\cot ^{3}(5x-1)(5x-1)'\frac{1}{\sin ^{2}(5x-1)}$
$=20\frac{\cos ^{3}(5x-1)}{\sin ^{5}(5x-1)}$
