http://www.ck12.org Chapter 2. Derivatives
7.x^2 y−y^2 x=0 at( 1 , 1 )- sin(xy) =yat(π 2 , 1 )
- Findy′′by implicit differentiation forx^3 y^3 =5.
- Use implicit differentiation to show that the tangent line to the curve 1 y^2 =kxat(x 0 ,y 0 )is given byyy 0 =
2 k(x+x^0 ), wherekis a constant.
d(y^2 )
dx =d(kx)
dx
2 ydydx=k
dy
dx=k
2 ySecond, we substitutey 0 fory, and that gives us the slopemof our tangent line at(x 0 ,y 0 ):
m= 2 ky 0Third, we set up the equation for our tangent line using point-slope form:
y−y 0 = 2 yk 0 (x−x 0 )Fourth, and finally, we manipulate this linear equation to get the termyy 0 isolated on the left hand side:
y−y 0 = 2 ky 0 (x−x 0 )
y= 2 ky 0 (x−x 0 )+y 0yy 0 = 2 k(x−x 0 )+(y 0 )^2
yy 0 = 2 k(x−x 0 )+kx 0 (Using the fact thaty^2 =kx)
yy 0 = 2 k(x+x 0 )