Derivatives of Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions

We will now begin to derive the derivatives of inverse trigonometric functions with basic trigonometry and Implicit Differentiation.

Theorem 1: The following functions have the following derivatives:
a) If $f(x) = \sin^{-1} x$, then $\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - x^2}}$.
b) If $f(x) = \cos ^{-1} x$, then $\frac{d}{dx} f(x) = - \frac{1}{\sqrt{1 - x^2}}$.
c) If $f(x) = \tan ^{-1} x$ then $\frac{d}{dx} f(x) = \frac{1}{1 + x^2}$.

We will now prove all parts of this property, all of which are similar.

  • Proof of a): First, let $y = \sin ^{-1} x$. It follows from the laws of inverse functions that:
\begin{align} \sin y = \sin(\sin ^{-1} x) \\ \sin y = x \end{align}
  • We can now use techniques of implicit differentiation in order to find the derivative of sin-1x.
\begin{align} \frac{d}{dx} \sin y = \frac{d}{dx} x \\ \frac{dy}{dx} \cos y = 1 \\ \frac{dy}{dx} = \frac{1}{\cos y} \\ \end{align}
  • Using a geometric argument, we can determine what $\cos y$ is in terms of $x$. We will first construct a triangle using the information given from $\sin y = x$:
  • Notice that this triangle describes $\sin y = x$ perfectly, from which we obtain that $\cos y = \sqrt{1 - x^2}$. Using this substitution into our derivative we get:
\begin{align} \frac{dy}{dx} = \frac{1}{\cos y} \\ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\\ \frac{d}{dx} \sin ^{-1}x = \frac{1}{\sqrt{1 - x^2}} \quad \blacksquare \end{align}
  • Proof of b): Let $y = \cos ^{-1} x$, and thus $\cos y = x$. Applying implicit differentiation techniques, we get:
\begin{align} \frac{d}{dx} \cos y = \frac{d}{dx} x \\ - \frac{dy}{dx} \sin y = 1 \\ \frac{dy}{dx} = - \frac{1}{\sin y} \end{align}
  • We will omit a visual representation of $\sin y = x$, however, acknowledge that the adjacent edge of the triangle will be equal to $x$, while the hypotenuse of the triangle will be $1$. Hence by the Pythagorean theorem, the opposite side to $y$ will be the square root of $\sqrt{1 - x^2}$. Thus, we obtain a value for $\sin y$, namely, $\sin y = \sqrt{1 - x^2}$, and by substitution into our derivative:
\begin{align} \frac{d}{dx} \cos ^{-1} x = - \frac{1}{\sqrt{1 - x^2}} \quad \blacksquare \end{align}
  • Proof of c): Let $y = \tan ^{-1} x$, and hence, $\tan y = x$. Using implicit differentiation techniques:
\begin{align} \frac{d}{dx} \tan y = \frac{d}{dx} x \\ \frac{dy}{dx} \sec ^2 y = 1 \\ \frac{dy}{dx} = \frac{1}{\sec^2y} \\ \frac{dy}{dx} = \cos^2 y \end{align}

Omitting the visual once again, we find the hypotenuse is equal to $\sqrt{1 + x^2}$ and furthermore, $\cos y = \frac{1}{\sqrt{1 + x^2}}$, so $\cos ^2 y = \frac{1}{1 + x^2}$. Making this substitution, we obtain that:

\begin{align} \frac{d}{dx} tan^{-1} x = \frac{1}{1 + x^2} \quad \blacksquare \end{align}

Example 1

Find the derivative of the function $f(x) = \ln x \cos ^{-1} x$.

We note that $f(x)$ has two main parts to it, so we will apply the product rule noting that $\frac{d}{dx} \ln x = \frac{1}{x}$ and $\frac{d}{dx} \cos ^{-1} x = - \frac{1}{\sqrt{1 - x^2}}$.

\begin{align} f(x) = \ln x \cos ^{-1} x \\ f'(x) = \ln x \left ( - \frac{1}{\sqrt{1 - x^2}} \right ) + \left ( \frac{1}{x} \right ) \cos^{-1} x \end{align}
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