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))/(2xy) then the equation of the curve is

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

The slope of the tangent to the curve y=6+x-x^(2) at (2,4) is

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

The slope of the tangent at a point P(x, y) on a curve is (- (y+3)/(x+2)) . If the curve passes through the origin, find the equation of the curve.

The slope of tangent at x=1 on the curve y=sin^(-1)(cos pi x) is

A curve passes through (1,(pi)/(4)) and at (x,y) its slope is (sin2y)/(x+tan y). Find the equation to the curve.

The slope of the tangent to the curve y = x + sin x cos x at x = (pi)/( 2) is