`x(dv)/dx=(1-2v-v^2)/(1+v)`

u=sin(m cos^(-1)x),v=cos(m sin^(-1)x), provethat (du)/(dv)=sqrt((1-u^(2))/(1-v^(2)))

If y=u//v," where "u,v," are differentiable functions of x and "vne0," then: "u(dv)/(dx)+v^(2)(dy)/(dx)=

int(1)/sqrt(1-2v^(2))dv

Prove that d/(dx)|(u_1,v_1,w_1),(u_2,v_2,w_2),(u_3,v_3,w_3)|=|(u_1,v_1,w_1),(u_2,v_2,w_2),(u_4,v_4,w_4)| where u,v,w are functions of x and (du)/(dx)=u_1,(d^2u)/(dx^2)=u_2, etc.

A body moves so that it follows the following relation (dv)/(dt)=-v^(2)+2v-1 where v is speed in m/s and t is time in second. If at t=0, v=0 then find the speed (in m/s) when acceleration in one fourth of its initial value.

int((1+v)/(1-v))dv

Evaluate : int ( (2v ) / ( 1 - v^2 ) ) dv

int(u(v(du)/(dx)-u(dv)/(dx)))/(v^(3))dx=

The Speed Of a particle moving in a plane is equal to the magnitude of its instantaneous velocity, v=|vl= sqrt(v_x^2+v_y^2). (a) Show that the rate of change of the speed is (dv)/(dt)=(v_xa_x+v_ya_y)/(sqrt(v_x^2+v_y^2)) . (b) Show that the rate of change of speed can be expressed as (dv)/(dt) is equal to a_t the component of a that is parallelto v.