$Z$ has a standard normal distribution. $P(Z \in (-a,b)) = 0.95$ where $a,b > 0$. Find the derivative the length, $l$, of the interval $(-a,b)$.

$Z$ has a standard normal distribution. $P(Z \in (-a,b)) = 0.95$ where $a,b > 0$. I'm asked to find the derivative the length, $l$, of the interval $(-a,b)$. I have an expression for $l$ as a function of $a$: $l = a + \Phi^{-1}(0.95 + \Phi(-a))$. How do I go about differentiating this? I am allowed to express it in terms of $\Phi, \Phi^{-1}$ and $\phi$. I assume some application of the chain rule, but I am a bit confused on how to go about this.

Thanks!

You can chain rule to differentiate inverse functions. Let $g(x) =f^{-1}(x)$. Then we have that

$f(g(x)) = x$. Differentiating both sides, we get that $g'(x)f'(g(x)) = 1$. Thus, we get that $g'(x) = \frac{1}{f'(g(x))}$. (Assuming the derivative is not zero, so we want functions without inflection points).

Thus in this case, we have that

$$l'(a) = 1 - (\Phi^{-1})'(0.95+\Phi(-a))\Phi'(-a) = 1 - \frac{\Phi'(-a)}{\Phi'(\Phi^{-1}(0.95+\Phi(-a)))} = 1 - \frac{\Phi'(-a)}{\Phi'(l(a)-a)}.$$