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Question Number 207127 by hardmath last updated on 07/May/24

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)={\sin }^{4}2\mathrm{x}$$$$\mathrm{Find}\mathrm{f}{}^{{}^{\prime }}\left(\frac{\pi }{12}\right)=?$$

Answered by Skabetix last updated on 07/May/24

$$\mathrm{f}\left(\mathrm{x}\right)={\sin }^4\left(2\mathrm{x}\right)$$$$\iff f^{\prime }\left(x\right)=8{sin}^3\left(2x\right)cos\left(2x\right)$$$$\mathrm{f}{}^{{}^{\prime }}\left(\frac{\pi }{12}\right)=8{sin}^3\left(\frac{\pi }{6}\right)cos\left(\frac{\pi }{6}\right)$$$$=8\times {\left(\frac{1}{2}\right)}^3\times \frac{\sqrt{3}}{2}$$$$=1\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$$$$$$

Commented by hardmath last updated on 07/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{ser}$$

Answered by mathzup last updated on 08/May/24

$$f^{{}^{\prime }}\left(x\right)=4\times 2cos\left(2x\right){sin}^3\left(2x\right)\Rightarrow$$$$f^{{}^{\prime }}\left(\frac{\pi }{12}\right)=8cos\left(\frac{\pi }{6}\right)\times {sin}^3\left(\frac{\pi }{6}\right)$$$$=8\times \frac{\sqrt{3}}{2}\times {\left(\frac{1}{2}\right)}^3=\frac{\sqrt{3}}{2}$$

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