Derivatives of Inverse Reciprocal Trigonometric Functions

# Derivatives of Inverse Reciprocal Trigonometric Functions

We will derive the derivatives of the inverse reciprocal trigonometric functions in the same manner that we derived the Derivatives of Inverse Trigonometric Functions by Implicit Differentiation.

 Theorem 1: The following functions have the following derivatives: a) If $f(x) = \csc ^{-1} x$, then $\frac{d}{dx} f(x) = - \frac{1}{| x |\sqrt{x^2 - 1}}$. b) If $f(x) = \sec ^{-1} x$, then $\frac{d}{dx} f(x) = \frac{1}{| x | \sqrt{x^2 - 1}}$. c) If $f(x) = \cot ^{-1} x$, then $\frac{d}{dx} f(x) = -\frac{1}{1 + x^2}$.
• Proof of a): Let $y = \csc^{-1} x$ so that $\csc y = x$. Applying implicit differentiation techniques:
(1)
\begin{align} \frac{d}{dx} \csc y = \frac{d}{dx} x \\ \frac{d}{dx} \frac{1}{\sin y} = \frac{d}{dx} x \\ \frac{0 \sin ^{-1} y - \frac{dy}{dx} \cos y}{\sin ^2y} = 1 \\ \frac{- \frac{dy}{dx} \cos y}{\sin ^2y} = 1 \\ - \frac{dy}{dx} = \frac{\sin ^2y}{\cos y} \\ \frac{dy}{dx} = - \frac{\sin ^2y}{\cos y} \\ \frac{dy}{dx} = - \tan y \sin y \end{align}
• We can construct a triangle depicting $\csc y = x$ and obtain that $\sin y = \frac{1}{\mid x \mid}$ and $\tan y = \frac{1}{\sqrt{x^2 - 1}}$.
 Note: We need to ensure that the derivative of cosecant inverse is negative because for the entire domain of cosecant inverse, the slopes are negative. There's a negative sign in our derivative thus far, that is $\frac{dy}{dx} = - \tan y \sin y$, so we need to ensure that both $\tan y$ and $\sin y$ are positive. Clearly, $\tan y = \frac{1}{\sqrt{x^2 - 1}}$ is positive, so to prevent problems, we will take the absolute value of $x$ such that $\sin y = \frac{1}{\mid x \mid}$. Note that our diagram does not suggest that $x$ can be negative since geometrically the length of the hypotenuse cannot be negative, though we insert this note here for clarification. • Substituting back into our formula for the derivative of cosecant we get that:
(2)
\begin{align} \frac{dy}{dx} = - \frac{1}{\sqrt{x^2 - 1}} \cdot \frac{1}{\mid x \mid} \\ \frac{dy}{dx} = - \frac{1}{\mid x \mid\sqrt{x^2 - 1}} \quad \blacksquare \end{align}
• Proof of b): Let $y = \sec ^{-1} x$ so that $\sec y = x$, and apply techniques of implicit differentiation to get that:
(3)
\begin{align} \frac{d}{dx} \sec y = \frac{d}{dx} x \\ \frac{dy}{dx} \sec y \tan y = 1 \\ \frac{dy}{dx} = \frac{1}{\sec y \tan y} \end{align}

Omitting a triangle construction to find the values of $\sec y$ and $\tan y$, we will note $\sec y = \mid x \mid$ and $\tan y = \sqrt{x^2 - 1}$.

 Note: For the same purposes as deriving the derivative of $\csc^{-1} x$, we should note that our derivative must always be positive since all slopes are positive. That can only happen provided $\sec y > 0$, and to eliminate this problem, we take the absolute value of $x$.
• We can substitute these values in to obtain our derivative:
(4)
\begin{align} \frac{dy}{dx} = \frac{1}{| x | \sqrt{x^2 - 1}} \quad \blacksquare \end{align}
• Proof of c): Let $y = \cot ^{-1} x$ so that $\cot y = x$, and applying implicit differentiation techniques we have that:
(5)
\begin{align} \frac{d}{dx} \cot y = \frac{d}{dx} x \\ \frac{dy}{dx} \left (-\csc ^2 y \right ) = 1 \\ \frac{dy}{dx} = \frac{1}{-\csc ^2 y} \end{align}
• Omitting a diagram, we find that $\csc y = \sqrt{1 + x^2}$ and thus $\csc ^2 y = 1 + x^2$. Substituting this into our derivative formula we get:
(6)
\begin{align} \frac{dy}{dx} = -\frac{1}{1 + x^2} \quad \blacksquare \end{align}