Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 159648 by cortano last updated on 19/Nov/21

$$y=\sin {}^2\left(2x\right)$$$$y^{\left(n\right)}=?$$

Answered by FongXD last updated on 19/Nov/21

$$\bullet \mathrm{y}=\frac{1}{2}(1-\cos 4\mathrm{x})=\frac{1}{2}-\frac{1}{2}\cos 4\mathrm{x}$$$$\bullet {\mathrm{y}}^{\prime }=2\sin 4\mathrm{x}=-2^{2\left(1\right)-1}\cos (\frac{\pi }{2}+4\mathrm{x})$$$$\bullet {\mathrm{y}}^{″}=8\sin (\frac{\pi }{2}+4\mathrm{x})=-2^{2\left(2\right)-1}\cos (\frac{2\pi }{2}+4\mathrm{x})$$$$\bullet {\mathrm{y}}^{‴}=32\sin (\frac{2\pi }{2}+4\mathrm{x})=-2^{2\left(3\right)-1}\cos (\frac{3\pi }{2}+4\mathrm{x})$$$$..........................................................$$$$\mathrm{suppose}\mathrm{that}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{true}\mathrm{all}\mathrm{the}\mathrm{way}\mathrm{up}\mathrm{to}\mathrm{k}-\mathrm{th}\mathrm{term}:{\mathrm{y}}^{\left(\mathrm{k}\right)}=-2^{2\mathrm{k}-1}\cos (\frac{\mathrm{k}\pi }{2}+4\mathrm{x})$$$$\mathrm{then}{\mathrm{y}}^{(\mathrm{k}+1)}={[-2^{2\mathrm{k}-1}\cos (\frac{\mathrm{k}\pi }{2}+4\mathrm{x})]}^{\prime }$$$$\iff {\mathrm{y}}^{(\mathrm{k}+1)}=2^{2\mathrm{k}-1}\times 4\sin (\frac{\mathrm{k}\pi }{2}+4\mathrm{x})$$$$\Rightarrow {\mathrm{y}}^{(\mathrm{k}+1)}=-2^{2(\mathrm{k}+1)-1}\cos [\frac{(\mathrm{k}+1)\pi }{2}+4\mathrm{x}]✓$$$$\mathrm{therefore},{\mathrm{y}}^{\left(\mathrm{n}\right)}=-2^{2\mathrm{n}-1}\cos (\frac{\mathrm{n}\pi }{2}+4\mathrm{x})$$

Commented by bobhans last updated on 19/Nov/21

$$\mathrm{or}{\mathrm{y}}^{\left(\mathrm{n}\right)}=-2^{2\mathrm{n}-1}\sin (4\mathrm{x}+\frac{\pi }{2}+\frac{\mathrm{n}\pi }{2})$$$${\mathrm{y}}^{\left(\mathrm{n}\right)}=-2^{2\mathrm{n}-1}\sin (4\mathrm{x}+(\mathrm{n}+1)\frac{\pi }{2})$$$$\mathrm{or}{\mathrm{y}}^{\left(\mathrm{n}\right)}=-2^{2\mathrm{n}-1}\sin (\frac{\pi }{2}-(4\mathrm{x}+\frac{\mathrm{n}\pi }{2}))$$$${\mathrm{y}}^{\left(\mathrm{n}\right)}=2^{2\mathrm{n}-1}\sin (4\mathrm{x}+(\mathrm{n}-1)\frac{\pi }{2})$$

Answered by mathmax by abdo last updated on 20/Nov/21

$$\mathrm{y}\left(\mathrm{x}\right)={\sin }^2\left(2\mathrm{x}\right)=\frac{1-\cos \left(4\mathrm{x}\right)}{2}=\frac{1}{2}-\frac{1}{2}\times \frac{{\mathrm{e}}^{\mathrm{i}4\mathrm{x}}+{\mathrm{e}}^{-\mathrm{i}4\mathrm{x}}}{2}$$$$=\frac{1}{2}-\frac{1}{4}{\mathrm{e}}^{4\mathrm{ix}}-\frac{1}{4}{\mathrm{e}}^{-4\mathrm{ix}}\Rightarrow \mathrm{for}\mathrm{n}>0$$$${\mathrm{y}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=-\frac{1}{4}{\left(4\mathrm{i}\right)}^{\mathrm{n}}{\mathrm{e}}^{4\mathrm{ix}}-\frac{1}{4}{(-4\mathrm{i})}^{\mathrm{n}}{\mathrm{e}}^{-4\mathrm{ix}}$$$$=-4^{\mathrm{n}-1}{\mathrm{i}}^{\mathrm{n}}{\mathrm{e}}^{4\mathrm{ix}}-4^{\mathrm{n}-1}{(-\mathrm{i})}^{\mathrm{n}}{\mathrm{e}}^{-4\mathrm{ix}}$$$$=-4^{\mathrm{n}-1}\{{\mathrm{i}}^{\mathrm{n}}{\mathrm{e}}^{4\mathrm{ix}}+{(-\mathrm{i})}^{\mathrm{n}}{\mathrm{e}}^{-4\mathrm{ix}}\}$$$$=-4^{\mathrm{n}-1}\{{\mathrm{e}}^{\frac{\mathrm{in}\pi }{2}}{\mathrm{e}}^{4\mathrm{ix}}+{\mathrm{e}}^{-\frac{\mathrm{in}\pi }{2}}{\mathrm{e}}^{-4\mathrm{ix}}\}$$$$=-4^{\mathrm{n}-1}\{{\mathrm{e}}^{\mathrm{i}(\frac{\mathrm{n}\pi }{2}+4\mathrm{x})}+{\mathrm{e}}^{-\mathrm{i}(\frac{\mathrm{n}\pi }{2}+4\mathrm{x})}\}$$$${\mathrm{y}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=-2.4^{\mathrm{n}-1}\cos (4\mathrm{x}+\frac{\mathrm{n}\pi }{2})$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com