If `x cosx dy/dx+(cosx-x sinx)y=1` the solution is `y=secx+c f(x),` then `f(x)=`

If y=(sinx)^(cosx) + (cos x)^sinx , then find dy/dx .

If y=(e^(2x)cosx)/(x sinx),then(dy)/(dx) =

If y=(e^(2x)cosx)/(x sinx),then(dy)/(dx) =

If int (e^(x)(1+sinx))/(1+cosx) dx = e^(x) f(x) + c then f(x) =

If f(x) = [(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)] then show f(x) . f(y) = f(x+y)

If f(x) = [(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)] then show f(x) . f(y) = f(x+y)

If y_(1)(x) and y_(2)(x) are two solutions of (dy)/(dx)+f(x)y=r(x), then y_(1)(x)+y_(2)(x) is solution of : (A) (dy)/(dx)+f(x)y=0 (B) (dy)/(dx)+2f(x)y=r(x)(C)(dy)/(dx)+f(x)y=2r(x)(D)(dy)/(dx)+2f(x)y=2r(x)

If int(1+cosx)/(cosx-cos^(2)x)dx=log|secx+tanx|-f(x)+C ,then f(x)=......

If y=(x)^(cosx)+(cosx)^(sinx) , find (dy)/(dx).

Find (dy)/(dx) for the functions: y=(x+sinx)/(x+cosx)