The slope of the tangent at `(x , y)` to a curve passing through `(1,pi/4)` is given by `y/x-cos^2(y/x),` then the equation of the curve is

The slope of the tangent at (x , y) to a curve passing through a point (2,1) is (x^2+y^2)/(2x y) , then the equation of the curve is

The slope of the tangent at (x , y) to a curve passing through a point (2,1) is (x^2+y^2)/(2x y) , then the equation of the curve is

The slope of the tangent to the curve y=sin^(-1) (sin x) " at " x=(3pi)/(4) is

A curve passes through (1,pi/4) and at (x,y) its slope is (sin 2y)/(x+tan y). Find the equation to the curve.

The slope of the tangent to the curve y=x^(3)+x+54 which also passes through the originis

Find the slopes of the tangent and the normal to the curve x y=6 at (1,\ 6)

The slope of the tangent to the curve y=cos^(-1)(cos x) " at " x=-(pi)/(4) , is

Find the slope of the tangent to the curve y = x^3- x at x = 2 .

The slope of the tangent to a curve at any point (x ,y) on its given by y/x-cot(y/x) . cos(y/x) ,(x >0,y >0) and the curve passes though the point (1,pi/4) . Find the equation of the curve.

A curve y=f(x) passes through point P(1,1) . The normal to the curve at P is a (y-1)+(x-1)=0 . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is