jsMath

Derivatives of inverse functions

       
Suppose f is a one-to-one function. For every point P1=(a,b)
on the graph of y=f(x), there is a corresponding point P2=(b,a) on the
graph of y=f1(x. The points P1 and P2 are reflections of each other
over the line y=x. In the illustration below, a=1.2 and b=3.32.
Now, suppose we have a line tangent to y=f(x) at the point (a,b). The
slope m of this line will be m=f(a). Since the graph of y=f1(x is
the reflection of the graph of y=f(x), the line tangent to y=f1(x
at (b,a) will have a slope of 1/m , the reciprocal of m.
Here is an animation showing the two tangent lines at various values of a:
       
Here is the graph of y=sinx, with domain restricted to [π/2,π/2] ,
and the inverse function y=arcsinx.
Here in an animation showing the lines tangent to y=sinx at (a,sina)
and y=arcsinx at (sina,a), for various values of a:
       
Here is a plot of f(x)=arcsinx, an animated tangent line,
and the derivative f(x)=11x2.
(The derivative may appear first.)