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},


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}.


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


\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“).

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