The answer is:
[tex](f^{-1})^{\prime}(1)=DNE[/tex]Explanation:
We need to use the theorem for the inverse of the derivative:
[tex](f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}[/tex]Then, we need to calculate the derivative of f and its inverse.
To find the derivative, we know that the derivative of cosine is negative sine, and by the chain rule:
[tex]f^{\prime}(x)=-2\sin(2x)[/tex]ANd to find the inverse, we call f(x) = y and solve for x:
[tex]\begin{gathered} y=\cos(2x) \\ \cos^{-1}(y)=2x \\ \end{gathered}[/tex][tex]x=\frac{\cos^{-1}(y)}{2}[/tex]Now, we switch the variables, and we have the inverse:
[tex]f^{-1}(x)=\frac{\cos^{-1}(x)}{2}[/tex]And now, by the theorem:
[tex](f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}[/tex]To find the derivative of the inverse in x = 1, we need to find first:
[tex]f^{-1}(1)=\frac{\cos^{-1}(1)}{2}=\frac{0}{2}=0[/tex]Now:
[tex](f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(0)}[/tex][tex]f^{\prime}(0)=-2\sin(2\cdot0)=-2\cdot0=0[/tex]And finally:
[tex](f^{-1})^{\prime}(x)=\frac{1}{0}=DNE[/tex]Thus, the correct answer is DNE
