If x=a(cos theta+log(tan theta/2)) and y = a sin theta , find d2y/dx2 .
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x = a(cos∅ +Iog(tan∅/2))
y = a sin∅
dx/d∅ = a( -sin∅ +1/tan∅/2 sec²∅/2.1/2)
= a{ -sin∅ + (1+tan²∅/2)/2tan∅/2}
=a( -sin∅ + cosec∅)
=a( 1-sin²∅)/sin∅
=acot∅.cos∅ --------(1)
dy/d∅ = acos∅ ------(2)
divide (2) /(1)
dy/dx = 1/cot∅
dy/dx = tan∅
differentiate wrt x
d²y/dx² = sec²∅.d∅/dx
d²y/dx² =sec²∅.{ 1/dx/d∅}
d²y/dx² = sec²∅.atan∅.sec∅
= a.sec³∅.tan∅
y = a sin∅
dx/d∅ = a( -sin∅ +1/tan∅/2 sec²∅/2.1/2)
= a{ -sin∅ + (1+tan²∅/2)/2tan∅/2}
=a( -sin∅ + cosec∅)
=a( 1-sin²∅)/sin∅
=acot∅.cos∅ --------(1)
dy/d∅ = acos∅ ------(2)
divide (2) /(1)
dy/dx = 1/cot∅
dy/dx = tan∅
differentiate wrt x
d²y/dx² = sec²∅.d∅/dx
d²y/dx² =sec²∅.{ 1/dx/d∅}
d²y/dx² = sec²∅.atan∅.sec∅
= a.sec³∅.tan∅
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