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Can anybody please help me with my work? I have to find the differentiate/ find the derivative of these two question: Please HELP!!!

  1. $sin^2(cos3x^3)^5 $
  2. $cot^2(x)((x^2)(3cos^3(3x)))^2$
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  • $\begingroup$ Chain rule chain rule chain rule chain rule $\endgroup$ – user137794 Apr 24 '14 at 20:33
  • $\begingroup$ i tried but i cannot get to the answer $\endgroup$ – Assab Amad Apr 24 '14 at 20:33
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    $\begingroup$ Why would a teacher assign such a ridiculous question? $\endgroup$ – user142299 Apr 24 '14 at 20:33
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    $\begingroup$ without proper parentheses, it is difficult to tell what is wanted, especially in the first question. $\endgroup$ – robjohn Apr 24 '14 at 20:52
  • $\begingroup$ I would propose to check this and this. You can create a free trial account to see the proof... $\endgroup$ – moray95 Apr 24 '14 at 21:37
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Chain rule

$$\begin{align} \frac{d}{dx}\sin^2(\cos3x^3)^5&=2\sin(\cos3x^3)^5\cdot\frac{d}{dx}\sin(\cos3x^3)^5\\ &=2\sin(\cos3x^3)^5\cdot\cos(\cos3x^3)^5\cdot\frac{d}{dx}(\cos3x^3)^5\\ &=2\sin(\cos3x^3)^5\cdot\cos(\cos3x^3)^5\cdot5(\cos3x^3)^4\cdot\frac{d}{dx}\cos3x^3\\ &=who\ the\ hell\ made\ you\ do\ this \end{align}$$

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    $\begingroup$ Run out of breath ? = $-2\sin(\cos3x^3)^5\cdot\cos(\cos3x^3)^5\cdot5(\cos3x^3)^4\cdot 9x^2\cdot\sin3x^3\\$ $\endgroup$ – Tom Collinge Apr 24 '14 at 20:47
  • $\begingroup$ The second question is the most difficult. $\endgroup$ – Assab Amad Apr 24 '14 at 21:22

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