Tangent lines? ?Find an equation of the line tangent to the following curves at the given point. y+ ?y = 6 ;(x,y) = (1,4)

Solution Step 1: Given curve is y + xy= 6 ;(x,y) = (1,4) The equation of tangent line to the curve y=f(x) at given point (x ,y ) with slope m is 0 0 given by (y y 0 = m(x x )0 Here (x ,0 )0(1,4) Now to find slope d (y + xy) = d (6) dx dx dy + d(xy) = 0 dx dx dy+ 1 d(xy) = 0 dx 2xy dx dy+ 1 (xdy + y (x)) = 0 dx 2xy dx dx dy 1 dy dx+ 2xy(xdx + y(1)) = 0 dy 1 dy dx+ 2xy(xdx + y) = 0 dy (1 + 2 xy dx + 2 xyy = 0 ( 2xy+1)dy = y 2xy dx 2xy dy y 2xy dx = 2xy (2xy+1 ) dy y dx = ( 2xy+1) Step 2 y Therefore slope m=( 2xy+1) Slope at point (1,4) is dy 4 4 4 dx(1,4)= m (1,4) 2(1)(4)+1 = 2 4+1 = 5