differentiate cos-1[(a+bcosx)/(b+acosx)] w r t x - Maths - Continuity and Differentiability

Question

differentiate cos-1[(a+bcosx)/(b+acosx)] w r t x

Devesh Kumar · Accepted Answer

y=cos-1 a+bcosxb+acosxdydx=dcos-1 a+bcosxb+acosxdxFormula to be used:dcos-1 xdx=-11-x2  & duvdx=vdudx-udvdxv2=-11-a+bcosxb+acosx2×b+acosx-bsinx-a+bcosx-asinxb+acosx2=-b+acosxb+acosx2-a+bcosx2×-b2sinx-abcosxsinx+a2sinx+absinxcosxb+acosx2=-b+acosxb2+a2cos2x+2abcosx-a2+b2cos2x+2abcosx×-b2sinx+a2sinxb+acosx2=-1b2+a2cos2x-a2+b2cos2x×a2-b2sinxb+acosx=-1a2-b2cos2x-a2-b2×a2-b2sinxb+acosx=1b2-a2×1-cos2x×b2-a2sinxb+acosx=b2-a2sinx×sinxb+acosx=b2-a2b+acosx  Answer

differentiate cos-1[(a+bcosx)/(b+acosx)] w.r.t x

differentiate cos^-1[(a+bcosx)/(b+acosx)] w.r.t x