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

Q. y = sin^(−1) (((a + bcosx)/(b + acosx))) , then find (dy/dx)

$$\boldsymbol{\mathrm{Q}}.\:\:\boldsymbol{{y}}\:=\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{a}}\:+\:\boldsymbol{{b}\mathrm{cos}{x}}}{\boldsymbol{{b}}\:+\:\boldsymbol{{a}\mathrm{cos}{x}}}\right)\:,\:\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}} \\ $$

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

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

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

. y = sin^(−1) (((a + bcosx)/(b + acosx))) ⇔siny=((a+bcosx)/(b+acosx)).Derivative two sides by x we get: y′cosy=((−bsinx(b+acosx)−(a+bcosx)(−asinx))/((b+acosx)^2 )) ⇔y′cosy=(((a^2 −b^2 )sinx)/((b+acosx)^2 )) ⇒y′=(((a^2 −b^2 )sinx)/((b+acosx)^2 cosy)) =(((a^2 −b^2 )sinx)/((b+acosx)^2 ))×(1/( (√(1−sin^2 y)))) (dy/dx)=(((a^2 −b^2 )sinx)/((b+acosx)^2 ))×(1/( (√(1−(((a+bcosx)/(b+acosx)))^2 )))) =(((a^2 −b^2 )sinx)/((b+acosx)^2 ))×∣((b+acosx)/( (√((b^2 −a^2 )sin^2 x))))∣

$$.\:\:\boldsymbol{{y}}\:=\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{a}}\:+\:\boldsymbol{{b}\mathrm{cos}{x}}}{\boldsymbol{{b}}\:+\:\boldsymbol{{a}\mathrm{cos}{x}}}\right) \\ $$$$\Leftrightarrow\mathrm{siny}=\frac{\mathrm{a}+\mathrm{bcosx}}{\mathrm{b}+\mathrm{acosx}}.\mathrm{Derivative}\:\mathrm{two}\:\mathrm{sides} \\ $$$$\mathrm{by}\:\mathrm{x}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{y}'\mathrm{cosy}=\frac{−\mathrm{bsinx}\left(\mathrm{b}+\mathrm{acosx}\right)−\left(\mathrm{a}+\mathrm{bcosx}\right)\left(−\mathrm{asinx}\right)}{\left(\mathrm{b}+\mathrm{acosx}\right)^{\mathrm{2}} } \\ $$$$\Leftrightarrow\mathrm{y}'\mathrm{cosy}=\frac{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{sinx}}{\left(\mathrm{b}+\mathrm{acosx}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{y}'=\frac{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{sinx}}{\left(\mathrm{b}+\mathrm{acosx}\right)^{\mathrm{2}} \:\mathrm{cosy}} \\ $$$$=\frac{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{sinx}}{\left(\mathrm{b}+\mathrm{acosx}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{y}}} \\ $$$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\frac{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{sinx}}}{\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{acosx}}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{bcosx}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{acosx}}}\right)^{\mathrm{2}} }}\: \\ $$$$\:=\frac{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{sinx}}}{\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{acosx}}\right)^{\mathrm{2}} }×\mid\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{acosx}}}{\:\sqrt{\left(\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}}\mid \\ $$

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