" 24."(dx)/(dt)+x cos t=(1)/(2)sin2t

dx/dt + x cos t = 1/2 sin 2 t

Find (dy)/(dx):x=a{cos t+(1)/(2)log tan^(2)(t)/(2)} and y=a sin t

If x=f(t), y=g(t) are differentiable functions of parameter 't' then prove that y is a differentiable function of 'x' and (dy)/(dx)=((dy)/(dt))/((dx)/(dt)),(dx)/(dt) ne 0 Hence find (dy)/(dx) if x=a cos t, y= a sin t.

Solve (dx)/(d t) = (t(2 log t + 1))/(sin x + x cos x)

If x=sin t-t cos t and y = t sin t +cos t, then what is (dy)/(dx) at point t=(pi)/(2)?

If x=a sin2t(1+cos2t) and y=b cos2t(1-cos2t), find the values of (dy)/(dx) at t=(pi)/(4) and t=(pi)/(3)

Find (dy)/(dx) when : x=a(2t+sin2t), y=a(1-cos2t)

If x = 2 cos t - cos 2 t and y =2 sin t - sin 2t, then (dy)/(dx) at t = (pi)/(2) is

Find (dy)/(dx) : x=a sin 2t (1+ cos 2t) and y= b cos 2t (1-cos 2t) show that, ((dy)/(dx))_(t = (pi)/(4))= (b)/(a)