Tangent lines? ?Find an equation of the line tangent to the following curves at the given point. x y+y = 75 ;(x,y) = (4,3)

Solution Step 1: 2 3 Given curve is x y + y = 75 ;(x,y) = (4,3) The equation of tangent line to the curve y=f(x) at given point (x ,y ) wi0h 0lope m is given by (y y ) = m(x x ) 0 0 Here (x ,0 )=04,3) Now to find slope d 2 3 d dx(x y + y ) = dx (75) d(x y) + d(y ) = 0 dx dx y d(x ) + x 2 d(y) + (3y ) dy = 0 ( d(uv)=u dv + v du ) dx dx dx dx dx dx y(2x) + x 2dy + (3y ) dy = 0 dx dx (2xy) + (x + 3y ) 2 dy = 0 dx (x + 3y )2 dy = (2xy) dx dy (2xy) dx= (x +3y ) Step 2 Therefore slope of the tangent line m= 22xy2 (x +3y ) Slope at point (4,3) is dy (2(4)(3)) = m (4,3)= 2 2 = 24 = 24 dx(4,3) ((4) +3(3) ) (16+27) 43