Question

Find (f^-1)'(0) for f(x)=cos2x on [0,pi/2].

Answer 1

hmm

Answer 2

\(\bf y=cos(2x)\qquad inverse\implies f^{-1}(x)=cos^{-1}(2x)
\\ \quad \\
then \qquad \cfrac{d}{dx}[cos^{-1}(2x)](0)\)

Answer 3

@jdoe0001 Do I multiply zero or plug in zero to the derivative of arccos2x?

Answer 4

plug in 0 for the derivative thus \(\bf (f^{-1})'(0)\)

Answer 5

plug in zero IN the derivative that is

Answer 6

Ok so \[(f^{-1})'(0)=-2\]?

Answer 7

@jdoe0001

Answer 8

\(\bf y=cos(2x)\qquad inverse\implies f^{-1}(x)=cos^{-1}(2x)
\\ \quad \\
then \qquad \cfrac{d}{dx}[cos^{-1}(2x)]\implies \cfrac{1}{\sqrt{1-(2x)^2}}
\\ \quad \\\
[cos^{-1}(2x)]'(0)\implies \cfrac{1}{\sqrt{1-(2(0))^2}}\)

Answer 9

should boil down to 1

Answer 10

@jdoe0001 Ok, thank you for your help.

Answer 11

ohh shoot.. one seec... should be negative...

Answer 12

\(\bf \cfrac{x^2 - 4}{2x^2}\\
y=cos(2x)\qquad inverse\implies f^{-1}(x)=cos^{-1}(2x)
\\ \quad \\
then \qquad \cfrac{d}{dx}[cos^{-1}(2x)]\implies -\cfrac{1}{\sqrt{1-(2x)^2}}
\\ \quad \\\
[cos^{-1}(2x)]'(0)\implies -\cfrac{1}{\sqrt{1-(2(0))^2}}\implies -1\)