What is the Derivative of arccot(x)?

The Mathematician — Fri, 23 Dec 2022 13:00:09 +0000

The derivative of

\text{arccot}(x)

-\frac{1}{x^2 + 1}

.

Proof. Let

F(x) = \text{arccot}(x) = \cot^{-1}(x)

. We have seen here that:

\begin{align*} \cot^{-1}(x) = \tan^{-1}(1/x), \quad x \neq 0. \end{align*}

So we have now

F(x) = \tan^{-1}(1/x)

. Take

f(u) = \tan^{-1}(u)

and

g(x) = \frac{1}{x}

such that

F(x) = f(g(x))

. We will be using the chain rule to find the derivative of

\tan^{-1}(1/x)

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We have seen here that

f'(u) = \frac{1}{1 + u^2}

and here that

g'(x) = -\frac{1}{x^2}

. So we have:

\begin{align*} f'(g(x)) = \frac{1}{1 + g(x)^2} = \frac{1}{1 + \frac{1}{x^2}}. \end{align*}

Substituting everything, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{1}{1 + \frac{1}{x^2}} \cdot \bigg(-\frac{1}{x^2}\bigg) \\ &= \frac{1}{\frac{x^2 + 1}{x^2}} \cdot \bigg(-\frac{1}{x^2}\bigg) \\ &= \frac{x^2}{x^2 + 1} \cdot \bigg(-\frac{1}{x^2}\bigg) \\ &= -\frac{x^2}{x^2(x^2 + 1)} \\ &= -\frac{1}{x^2 + 1}. \end{align*}

So, the derivative of

\text{arccot}(x)

-\frac{1}{x^2 + 1}

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