Derivative of inverse trig functions

Above is a link to the question! Please correct me if I'm wrong but, I'm guessing that the tangents lines at +-1 of the graph do not exist because the tangent is a vertical line which is undefined.

I feel like i'm supposed to solve this quesiton using Mean Value Theorem...

• Yes, $\tan(x)=\pm1$ is undefined because the tangent would just be a horizontal line that never intersects the x-axis. – Badr B Nov 29 '17 at 23:52

You hit it right on the head. As $|x|$ approaches $1$, the tangent line's slope approaches infinity because of the zero in the denominator.
A simple example is $f(x) = x^2$ on $[0, 1]$. Its slope at $0$ is $0$, so the inverse function, $\sqrt{x}$, has infinite slope there.
• How is there an infinite slope from [0,1] in $\sqrt{x}$? It doesn't go to infinity if it's a closed interval rgiht?... – user13123 Nov 30 '17 at 5:12
• The slope at 0 goes to $\infty$. $(\sqrt{x})' = (x^{1/2})' = (1/2)x^{-1/2}$. – marty cohen Nov 30 '17 at 7:24