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Question Number 146899 by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)={\sin }^5\mathrm{x}\mathrm{calculate}{\mathrm{f}}^{\left(5\right)}\left(\frac{\pi }{2}\right)$$

Answered by Olaf_Thorendsen last updated on 17/Jul/21

$$f\left(x\right)={\sin }^{5}x$$$$f\left(x\right)={\left(\frac{e^{ix}-e^{ix}}{2i}\right)}^5$$$$f\left(x\right)=\frac{1}{16\times 2i}(e^{5ix}-5e^{3ix}+10e^{ix}-10e^{-ix}+5e^{-3ix}-e^{-5ix})$$$$f\left(x\right)=\frac{\sin 5x-5\sin 3x+10\sin x}{16}$$$$f^{\left(5\right)}\left(x\right)=\frac{5^5\cos 5x-5.3^5\cos 3x+10\cos x}{16}$$$$f^{\left(5\right)}\left(\frac{\pi }{2}\right)=0$$

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