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Question Number 106847 by mathocean1 last updated on 07/Aug/20

$$Givenf\left(x\right)=\frac{3\sqrt{3}}{sinx}+\frac{1}{cosx}$$$$showthatf{}^{\prime }\left(x\right)=cosx\frac{({tan}^{3}x-3\sqrt{3})}{{sin}^{2}x}$$

Answered by bemath last updated on 07/Aug/20

$${}^{@\mathrm{bemath}@}$$$$\mathrm{f}\left(\mathrm{x}\right)=3\sqrt{3}\csc \mathrm{x}+\sec \mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-3\sqrt{3}\cot \mathrm{x}\csc \mathrm{x}+\sec \mathrm{x}\tan \mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{\sin \mathrm{x}}{\cos {}^2\mathrm{x}}-\frac{3\sqrt{3}\cos \mathrm{x}}{\sin {}^2\mathrm{x}}$$$$=\frac{\sin {}^3\mathrm{x}-3\sqrt{3}\cos {}^3\mathrm{x}}{\cos {}^2\mathrm{x}\sin {}^2\mathrm{x}}$$$$=\frac{\tan {}^3\mathrm{x}-3\sqrt{3}}{\sin {}^2\mathrm{x}\left(\frac{1}{\cos \mathrm{x}}\right)}=\frac{\cos \mathrm{x}(\tan {}^3\mathrm{x}-3\sqrt{3})}{\sin {}^2\mathrm{x}}$$

Commented by mathocean1 last updated on 07/Aug/20

$$Pleasesircanyoudetailthelast$$$$line?I^{\prime }mnotunderstandinghow$$$${tan}^{3}appeared...$$

Commented by bemath last updated on 07/Aug/20

$$\mathrm{from}\frac{\sin {}^3\mathrm{x}-3\sqrt{3}\cos {}^3\mathrm{x}}{\cos {}^2\mathrm{x}\sin {}^2\mathrm{x}}:\frac{\cos {}^3\mathrm{x}}{\cos {}^3\mathrm{x}}$$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

$$f\left(x\right)=\frac{3\sqrt{3}}{sinx}+\frac{1}{cosx}$$$$f^{\prime }\left(x\right)=\frac{-3\sqrt{3}cosx}{{sin}^{2}x}+secxtanx=\frac{-3\sqrt{3}cosx}{{sin}^{2}x}+\frac{sinx}{{cos}^{2}x}$$$$=\frac{-3\sqrt{3}{cos}^{3}x+{sin}^{3}x}{{sin}^2{xcos}^{2}x}=\frac{-3\sqrt{3}+\frac{{sin}^{3}x}{{cos}^{3}x}}{{sin}^{2}x\frac{{cos}^{2}x}{{cos}^{3}x}}=cosx\left(\frac{{tan}^{3}x-3\sqrt{3}}{{sin}^{2}x}\right)$$

Commented by bemath last updated on 07/Aug/20

$$\mathrm{typo}\mathrm{sir}.\mathrm{it}\tan {}^3\mathrm{x}$$

Answered by 1549442205PVT last updated on 07/Aug/20

$$f\left(x\right)=\frac{3\sqrt{3}}{sinx}+\frac{1}{cosx}$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-\frac{3\sqrt{3}}{{\sin }^2\mathrm{x}}\times \mathrm{cosx}+\frac{-1}{{\cos }^2\mathrm{x}}\times (-\mathrm{sinx})$$$$(\mathrm{since}{\left(\frac{1}{\mathrm{u}}\right)}^{\prime }=\frac{-1}{{\mathrm{u}}^2}\times {\mathrm{u}}^{\prime })$$$$=\frac{-3\sqrt{3}\mathrm{cosx}}{{\sin }^2\mathrm{x}}+\frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}=\frac{-3\sqrt{3}\mathrm{cosx}}{{\sin }^2\mathrm{x}}+\frac{{\sin }^3\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}$$$$=\frac{-3\sqrt{3}\mathrm{cosx}}{{\sin }^2\mathrm{x}}+\frac{\left(\frac{{\sin }^3\mathrm{x}}{{\cos }^3\mathrm{x}}\right)\mathrm{cosx}}{{\sin }^2\mathrm{x}}$$$$=\frac{-3\sqrt{3}\mathrm{cosx}}{{\sin }^2\mathrm{x}}+\frac{{\tan }^3\mathrm{x}.\mathrm{cosx}}{{\sin }^2\mathrm{x}}$$$$=\mathbf{cosx}\left(\frac{{\mathbf{tan}}^3\mathbf{x}-3\sqrt{3}}{{\mathbf{sin}}^2\mathbf{x}}\right)(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

Commented by mathocean1 last updated on 08/Aug/20

$$Thankyousirs...$$

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