Two Calculus Problems

Question

Question 1.

Suppose s(t) = cos(t)+t.  Note (0,1) is a point on the graph of s.  Compute (s^-1)'(1).

 

Question 2.

Suppose f(1)=2, f'(1)=-3, f(2)=7, and f'(2)=4.  Set F(x) = f(f(x)).  What is F'(1)?  

Answer 1

For Question 1: Since function s is monotonic and continuous on [0,pi]... if c is a value of t on the interval cited above, then s has an inverse and the following is true: if f(c) = d, then (s inverse)'(d) = 1/s'(c). s'(t) = -sin t + 1 s(0) = 1 s'(0) = 1 so c=0, d=1 Therefore (s inverse)'(1) = 1/s'(0) = 1/1 = 1.