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Question Number 112714 by deepak@7237 last updated on 09/Sep/20

$$\mathbf{Q}.\mathit{y}={\mathbf{sin}}^{-1}\left(\frac{\mathit{a}+\mathit{b}\mathbf{cos}\mathit{x}}{\mathit{b}+\mathit{a}\mathbf{cos}\mathit{x}}\right),\mathbf{then}\mathbf{find}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$$

Answered by deepak@7237 last updated on 09/Sep/20

Commented by deepak@7237 last updated on 09/Sep/20

Answered by 1549442205PVT last updated on 09/Sep/20

$$.\mathit{y}={\mathbf{sin}}^{-1}\left(\frac{\mathit{a}+\mathit{b}\mathbf{cos}\mathit{x}}{\mathit{b}+\mathit{a}\mathbf{cos}\mathit{x}}\right)$$$$\iff \mathrm{siny}=\frac{\mathrm{a}+\mathrm{bcosx}}{\mathrm{b}+\mathrm{acosx}}.\mathrm{Derivative}\mathrm{two}\mathrm{sides}$$$$\mathrm{by}\mathrm{x}\mathrm{we}\mathrm{get}:$$$${\mathrm{y}}^{\prime }\mathrm{cosy}=\frac{-\mathrm{bsinx}(\mathrm{b}+\mathrm{acosx})-(\mathrm{a}+\mathrm{bcosx})(-\mathrm{asinx})}{{(\mathrm{b}+\mathrm{acosx})}^2}$$$$\iff {\mathrm{y}}^{\prime }\mathrm{cosy}=\frac{({\mathrm{a}}^2-{\mathrm{b}}^2)\mathrm{sinx}}{{(\mathrm{b}+\mathrm{acosx})}^2}$$$$\Rightarrow {\mathrm{y}}^{\prime }=\frac{({\mathrm{a}}^2-{\mathrm{b}}^2)\mathrm{sinx}}{{(\mathrm{b}+\mathrm{acosx})}^2\mathrm{cosy}}$$$$=\frac{({\mathrm{a}}^2-{\mathrm{b}}^2)\mathrm{sinx}}{{(\mathrm{b}+\mathrm{acosx})}^2}\times \frac{1}{\sqrt{1-{\sin }^2\mathrm{y}}}$$$$\frac{\mathbf{dy}}{\mathbf{dx}}=\frac{({\mathbf{a}}^2-{\mathbf{b}}^2)\mathbf{sinx}}{{(\mathbf{b}+\mathbf{acosx})}^2}\times \frac{1}{\sqrt{1-{\left(\frac{\mathbf{a}+\mathbf{bcosx}}{\mathbf{b}+\mathbf{acosx}}\right)}^2}}$$$$=\frac{({\mathbf{a}}^2-{\mathbf{b}}^2)\mathbf{sinx}}{{(\mathbf{b}+\mathbf{acosx})}^2}\times \mid \frac{\mathbf{b}+\mathbf{acosx}}{\sqrt{({\mathbf{b}}^2-{\mathbf{a}}^2){\mathbf{sin}}^2\mathbf{x}}}\mid$$

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