Implicit Differentiation

Clear[x,y, lhs, rhs]

Consider the implicitly defined curve  (x^2)+xy-(y^3)=12+sin(x+y).  Let us find the tangent line to this curve at the point P=(1,-1)

ContourPlot[(x^2)+x y -(y^3)==Cos[x+y],{x,-2,5}, {y,-2,5},ContourStyle->{Blue, Thickness[0.01]}]

lhs = D[(x^2)+x y[x] -((y[x])^3), x]

rhs=D[Cos[x+y[x]], x]

Solve[lhs==rhs, y'[x]]

Thus, in standard notation, dy/dx is given by the expression:

-((2x+sin(x+y)+y)/(x+sin(x+y)-3 (y^2)))

Next, let us find the tangent line to the Astroid, (x^(2/3))+(y^(2/3))  =5, at the  point P = (8,1)

Clear[x, y, lhs, rhs, m]

ContourPlot[  ((Abs[x])^(2/3))+((Abs[y])^(2/3)) == 5,{x,-12,12}, {y,-12,12}]

lhs = D[ (x^(2/3))+((y[x])^(2/3)), x]

rhs=D[5,x]

Solve[lhs==rhs, y'[x]]

To evaluate dy/dx at the point (8, 1) we define a function, h, of two variables

h[x_, y_]=-((y^(1/3))/(x^(1/3)))

m=h[8,1]

So the equation of the tangent line to our implicitly defined curve at P=(8,1) is given by:  y-1 = m (x-8)

Now we will graph the Astroid together with its tangent and normal lines at the point P = (8,1)

ContourPlot[  {((Abs[x])^(2/3))+((Abs[y])^(2/3)) == 5,y-1==m(x-8), y-1==(-1/m) (x-8)},{x,-12,15}, {y,-12,12}, AspectRatio->Automatic, ContourStyle->{{Blue, Thickness[0.01]}, Red, Green}]

Below are more examples of plots of implicitly defined functions.

ContourPlot[{((((x^2)+(y^2)))^2)==(x^2)-(y^2),((((x^2)+(y^2)))^2)==2 x y},{x,-1,1},{y,-1,1},ContourStyle->{Blue,{Red,Dashing[{.05}]}}]

ContourPlot[{((((x^2)+(y^2)))^2)==(x^2)-(y^2),(x^2)+(y^2)==1},{x,-1,1},{y,-1,1},ContourStyle->{GrayLevel[0],Dashing[{.03}]}]

Parametric Equations

Consider the Involute of a circle, defined by x(t) = cos t + t sin t,  y(t) = sin t - t cos t, where  0 ⩽ t ⩽ 10.

ParametricPlot[{Cos[t] + t Sin[t], Sin[t]-t  Cos[t]}, {t, 0, 10}]

To find the slope of the curve as a function of t, we use the fact that dy/dx = (dy/dt)/(dx/dt), which follows from the Chain Rule.

Clear[f, g, slope]

f[t_] := Cos[t] + t Sin[t]

g[t_] := Sin[t]-t  Cos[t]

slope = ((D[g[t], t])/(D[f[t], t]))

To plot the equation of the tangent line (y = mx+b) to the Involute of a circle at t = 2, we write the line in parametric form setting  x(t) = t.   
For the involute of a circle at t = 2:  x = t, y = tan[2] (t -x[2]) + y(2)

ParametricPlot[{{Cos[t] + t Sin[t], Sin[t]-t  Cos[t]}, {t, Tan[2] (t - f[2]) + g[2]}}, {t, 0, 10}]

Next, consider the Teardrop, defined by x(t) = 2  cos t -  sin 2t,  y(t) = 4 sin t

Clear[t, f, g, slope]

ParametricPlot[{2 Cos[t] - Sin[2 t], 4 Sin[t]}, {t, 0, 2 Pi}]

Here again, the slope of the curve at t  is given by (dy/dt)/(dx/dt):

f[t_] = 2 Cos[t] - Sin[2 t]

g[t_]=4 Sin[t]

slope = ((D[g[t], t])/(D[f[t], t]))

Just for fun, here is a "neat example" of a parametric curve given in Mathematica:

ParametricPlot[{Sin[u]u ,Cos[u]u}, 
{u,0,100},PlotPoints->125,Axes->False,ColorFunction->(ColorData["Rainbow"][#3]&),MaxRecursion->0,PlotStyle->Thick]