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Question Number 122216 by help last updated on 14/Nov/20

Answered by liberty last updated on 14/Nov/20

$$\ell \mathrm{n}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arc}\sin 4\mathrm{x}$$$$\frac{\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{4}{\sqrt{1+{16\mathrm{x}}^2}}\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{{4\mathrm{e}}^{\mathrm{arc}\sin 4\mathrm{x}}}{\sqrt{1+{16\mathrm{x}}^2}}$$$$\mathrm{f}{}^{\prime }\left(\frac{1}{8}\right)=\frac{{4\mathrm{e}}^{\frac{\pi }{6}}}{\sqrt{1+\frac{1}{4}}}=\frac{{8\mathrm{e}}^{\frac{\pi }{6}}}{\sqrt{5}}.▴$$

Commented by help last updated on 14/Nov/20

$$laststepplease.$$$$explain$$

Commented by liberty last updated on 15/Nov/20

$$\mathrm{f}{}^{\prime }\left(\frac{1}{8}\right)=\frac{{4\mathrm{e}}^{\mathrm{arc}\sin \left(\frac{4}{8}\right)}}{\sqrt{1+\frac{16}{64}}}=\frac{{4\mathrm{e}}^{\mathrm{arc}\sin \left(\frac{1}{2}\right)}}{\sqrt{1+\frac{1}{4}}}$$$$=\frac{{4\mathrm{e}}^{\frac{\pi }{6}}}{\sqrt{\frac{5}{4}}}=\frac{{4\mathrm{e}}^{\frac{\pi }{6}}}{\frac{1}{2}\sqrt{5}}=\frac{{8\mathrm{e}}^{\frac{\pi }{6}}}{\sqrt{5}}$$

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