Question

x=sin^{3} t / \sqrt{cos 2t}, y = cos^{3}t / \sqrt cos 2t find dy/dx

Answer 1

dy/dx = - cot3t या dy/dx =  Cot²t( Tan2t - 3Tant)/ ( 3  + TantTan2t) , x = Sin³t/√Cos2t , y = Cos³t/√Cos2t

Step-by-step explanation:

dy/dx

x = Sin³t/√Cos2t

y = Cos³t/√Cos2t

x = Sin³t/√Cos2t

dx/dt  =  3Sin²tCost/√Cos2t   + Sin³t(-1/2Cos2t√Cos2t)(-2Sin2t)

=> dx/dt  = ( 3Sin²tCost  + Sin³tTan2t)/√Cos2t)

=> dx/dt  =Sin²tCost ( 3  + TantTan2t)/√Cos2t)

y = Cos³t/√Cos2t

dy/dt  =  3Cos²t(-Sint)/√Cos2t   + Cos³t(-1/2Cos2t√Cos2t)(-2Sin2t)

dy/dt  =Cos³t( -3Tant  + Tan2t)/√Cos2t)

(dy/dt)/(dx/dt)  =  Cos³t( -3Tant  + Tan2t)/Sin²tCost ( 3  + TantTan2t)

=> dy/dx =  Cot²t( Tan2t - 3Tant)/ ( 3  + TantTan2t)

Tan2t = 2Tant/(1 - Tan²t)

dy/dx = - cot3t

और अधिक जानें :

sin(x²+5)"

brainly.in/question/15286193

sin (ax+b) फलन का अवकलन कीजिए

brainly.in/question/15286166