(*Video 1: Implicit differentiation *)

f[x]^4(*f is an unspecified function of x*)

D[f[x]^4,x](*we can differentiate f^4 without knowing what f is:*)

f[x_]:=2x+3

Expand[4 ((f[x])^3) f'[x]]

Expand[D[f[x]^4,x]]

Clear[f]

(*the same logic applies to more complicated examples: *)
D[f[x]^4+2x f[x]+Cos[2f[x]]+x^3,x]

D[3x^2+7y[x]^2,x](* y is a function of x'*)

D[a[x]b[x]+a[x]^2+E^(b[x]),x](*a and b are functions of x*)

(*Video 2: Application of implicit differentiation*)

ellipse=ContourPlot[3x^2+7y^2==1,{x,-1,1},{y,-1,1}]

(*Find the tangent line to the ellipse above at the point {x->((Sqrt[3])/4),y->(1/4)} *)

3x^2+7y^2/.{x->((Sqrt[3])/4),y->(1/4)}

D[3x^2+7y[x]^2,x]==0

Solve[6 x+14 y[x] y'[x]==0,y'[x]]

(*Slope of the tangent line at the given point:*)
m=-((3 x)/(7 y))/.{x->((Sqrt[3])/4),y->(1/4)}

(*Tangent line to graph of y[x] at (x0,y0) is given by:*)
Collect[y0+m(x-x0)/.{x0->((Sqrt[3])/4),y0->(1/4)},x]

line=Plot[(4/7)-((3 (Sqrt[3]) x)/7),{x,-3,3},PlotStyle->Gray];
Show[ellipse,line]

(*Can also do it explicitly:*)
Solve[3x^2+7y^2==1,y]

Plot[((Sqrt[1-3 (x^2)])/(Sqrt[7])),{x,-1,1},PlotRange->{{-1,1},{-1,1}}]

D[((Sqrt[1-3 (x^2)])/(Sqrt[7])),x]

(*So we find, once again, the slope of the tangent line to be:*)
-((3 x)/((Sqrt[7]) (Sqrt[1-3 (x^2)])))/.x->((Sqrt[3])/4)

m (*agrees with earlier computation!*)