tính đạo hàm hs sau 1) y=tan^4 (pi/6+2x) 2) y=cot^4 (5x-1)

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1. 1.

   $y’=4\tan^3\Big(\dfrac{\pi}{6}+2x\Big).\Big[ \tan\Big(\dfrac{\pi}{6}+2x\Big)\Big]’$

   $=\dfrac{\tan^3\Big(\dfrac{\pi}{6}+2x\Big)}{2\cos^2\Big(2x+\dfrac{\pi}{6}\Big)}$

   2.

   $y’=4\cot^3(5x-1).[\cot(5x-1)]’$

   $=\dfrac{-5\cot^3(5x-1)}{4\sin^2(5x-1)}$

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2. 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)
   $y’=4\cot ^{3}(5x-1)\cot ‘(5x-1)$
   $=4\cot ^{3}(5x-1)(5x-1)’\frac{1}{\sin ^{2}(5x-1)}$
   $=20\frac{\cos ^{3}(5x-1)}{\sin ^{5}(5x-1)}$

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