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Question Number 114066 by john santu last updated on 17/Sep/20

$$Givenf\left(x\right)=7+\cos 2x+2\sin {}^{2}x$$$$find{f}^{\left(10\right)}\left(x\right)?$$

Commented by bemath last updated on 17/Sep/20

$$f\left(x\right)=7+1-2\sin {}^{2}x+2\sin {}^{2}x$$$$f\left(x\right)=8\Rightarrow f{}^{\left(10\right)}\left(x\right)=0$$

Answered by Olaf last updated on 17/Sep/20

$$f\left(x\right)=7+\cos 2x+{2\sin }^{2}x$$$$f\left(x\right)=7+\cos 2x+2\frac{1-\cos 2x}{2}$$$$f\left(x\right)=8$$$$f^{\left(k\right)}\left(x\right)=0$$

Answered by 1549442205PVT last updated on 17/Sep/20

$$\mathrm{Since}\cos 2\mathrm{x}=1-{2\sin }^2\mathrm{x},\cos 2\mathrm{x}+{2\sin }^2\mathrm{x}$$$$=1\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=7+\cos 2\mathrm{x}+{2\sin }^2\mathrm{x}=8$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=0\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=0\mathrm{\forall }\mathrm{n}⩾1,\mathrm{n}\in \mathrm{N}$$

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