# Finding Derivative the “Hard” Way

In “Instrumental Flying“, we defined $\cosh^{-1}(x), \sinh^{-1}(x)$ as the inverse of $\cosh(x)$ and $\sinh(x)$ repectively.

To find the derivative of $\cosh^{-1}(x)$, let

$y = \cosh^{-1}(x).$

We have

$x = \cosh(y).$

Differentiate it,

$\frac{d}{dx} x = \frac{d}{dx} \cosh(y) \implies 1=\frac{d}{dy} \cosh(y)\cdot \frac{dy}{dx},$

i.e.,

$1 = \sinh(y) \frac{dy}{dx}\implies \frac{dy}{dx} = \frac{1}{\sinh(y)}.$

By $\cosh(y)^2-\sinh(y)^2=1$ (see Exercise-1) and $\cosh(y) \ge 1$ (see Exrecise-2),

$\sinh(y)^2 = \cosh(y)^2-1 \implies |\sinh(y)| = \sqrt{x^2-1} \overset{ (\star) }{\implies} \sinh(y) = \sqrt{x^2-1}.$

And so,

$\frac{dy}{dx} = \frac{1}{\sqrt{x^2-1}} \implies \boxed{\frac{d}{dx}\cosh^{-1}(x) = \frac{1}{\sqrt{x^2-1}}}.$

Similarly, to find $\frac{d}{dx}\sinh^{-1}(x),$ let

$y = \sinh^{-1}(x)\implies x=\sinh(y).$

Differentiate it,

$\frac{d}{dx} x = \frac{d}{dx}\sinh(y) = \frac{d}{dy}\sinh(y)\frac{dy}{dx}\implies 1 = \cosh(y)\cdot\frac{dy}{dx}.$

By $\cosh(x)^2-\sinh(x)^2=1, \cosh(y) \ge 1$,

$\cosh(y) = \textbf{+}\sqrt{\sinh(y)^2+1} = \sqrt{x^2+1}.$

Therefore,

$1 = \sqrt{x^2+1}\frac{dy}{dx} \implies \boxed{\frac{d}{dx}\sinh^{-1}(x) = \frac{1}{\sqrt{x^2+1}}}$.

Prove:

$\forall x \ge 0, \sinh(x) \ge 0.\quad\quad\quad(\star)$

By definition,

$\sinh(x) = \frac{e^x-e^{-x}}{2} = \frac{e^{2x}-1}{2 e^{x}} \ge 0$, since $\forall x>0, e^{x},e^{2x} \ge 1$ (see Exercise-3).

Exercise-1 Show that $\forall x \in R, \cosh(x)^2 - \sinh(x)^2 =1$.

Exercise-2 Show that $\forall x \in R, \cosh(x) \ge 1$.

Exercise-3 Show that $\forall x \ge 0, e^{x} \ge 1.$

Exercise-4 Differentiate $\cos^{-1}(x), \sinh^{-1}(x)$ directly (hint: see”Deriving Two Inverse Functions“).