`d/dx(cosx.sin^-1x)`

d/dx((1-cosx)/sinx)=

d/dx((1-cosx)/sinx)=

The differentiation of cosx with respect to x' is -sinxdot i.e. d/(dx)(cosx)=-sinx

Let f(x) = {(sin (tan^(-1) x) + sin (cot^(-1) x)}^2 - 1 , where |x| gt 1 and dy/dx = 1/2 d/dx (sin^(-1) f(x)) . If y(sqrt3) = pi/6 then y( -sqrt3)

d/(dx) (sin^(-1) "" (2x)/(1+x^(2))) is equal to

Solve the differential equation (dy)/(dx)=sin^-1x

Find the general solution of the differential equations (dy)/(dx)=sin^(-1)x

Differentiate the following function with respect of x :(sinx+cosx)/(sinx-cosx)

Differentiate tan^(-1){(cosx+sinx)/(cosx-sinx)} with respect to x

Differentiate with respect to x, (x^(5)-cosx)/(sinx)