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 32971 by math1967 last updated on 08/Apr/18

If siny=xsin(a+y) show that  (dy/dx)=((sina)/(1−2xcosa+x^2 ))

$${If}\:{siny}={xsin}\left({a}+{y}\right)\:{show}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\frac{{sina}}{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$

Answered by math1967 last updated on 06/May/18

siny=xsinacosy +xcosasiny  ⇒siny(1−xcosa)=xsinacosy  ⇒tany=((xsina)/((1−xcosa)))  ⇒sec^2 y(dy/dx) =((sina)/((1−xcosa)^2 ))  ⇒(dy/dx)=((sina)/((1−xcosa)^2 ))×(1/(1+tan^2 y))  (dy/dx)=((sina)/((1−xcosa)^2 ))×(1/(1+((x^2 sin^2 a)/((1−xcosa)^2 ))))  ⇒(dy/dx)=((sina)/((1−xcosa)^2 ))×(((1−xcosa)^2 )/(1−2xcosa+x^2 (sin^2 a+cos^2 a)))  =((sina)/(1−2xcosa+x^2 ))

$${siny}={xsinacosy}\:+{xcosasiny} \\ $$$$\Rightarrow{siny}\left(\mathrm{1}−{xcosa}\right)={xsinacosy} \\ $$$$\Rightarrow{tany}=\frac{{xsina}}{\left(\mathrm{1}−{xcosa}\right)} \\ $$$$\Rightarrow{sec}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}\:=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {y}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {a}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} \left({sin}^{\mathrm{2}} {a}+{cos}^{\mathrm{2}} {a}\right)} \\ $$$$=\frac{{sina}}{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} } \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com