Note that arc(cotx) is an inverse trigonometric function. We will find the derivative of arc(cotx) and it is equal to -1/(1+x^{2}). In this post, we will learn how to differentiate the function arc cotx.

**What is the derivative of arc cotx?**

**Answer: **The derivative of arccotx is -1/(1+x^{2}).** **

**Solution:**

**[Method 1]**:

Let y = arc cotx …(∗)

As we need to find the derivative of arccot x, so we have to calculate dy/dx.

From (∗) we get that

y = cot^{-1}x

⇒ coty = x …(∗∗)

We now differentiate both sides of (∗∗) with respect to x. So we obatin that

$\dfrac{d}{dx}$(coty) = $\dfrac{dx}{dx}$

⇒ – cosec^{2}y $\dfrac{dy}{dx}$ =1

⇒ $\dfrac{dy}{dx}$ = -1/cosec^{2}y

⇒ $\dfrac{dy}{dx}$ = – $\dfrac{1}{1+\cot^2 y}$ as we know that cosec^{2}y = 1+cot^{2}y

⇒ $\dfrac{dy}{dx}$ = – $\dfrac{1}{1+x^2}$ as coty = x by (∗∗)

Thus, the derivative of arc cotx, that is, the differentiation of cot inverse x is equal to – 1/(1+x^{2}).

**[Method 2]**:

(Using the derivative of tan inverse x)

We know that

tan^{-1}x + cot^{-1}x = π/2

Differentiating both sides of the above equation, we get that

$\dfrac{d}{dx}$(tan^{-1}x) + $\dfrac{d}{dx}$(cot^{-1}x) = 0 as we know that the derivative of a constant is zero and π/2 here is a constant.

⇒ 1/(1+x^{2}) + $\dfrac{d}{dx}$(cot^{-1}x) =0

⇒ $\dfrac{d}{dx}$(cot^{-1}x) = -1/(1+x^{2}).

Thus, we have proven that the derivative of arccot x is -1/(1+x^{2}).

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## FAQs

### Q1: If y=arc cotx, then find dy/dx?

Answer: If y=arc cotx, then differentiating both sides we get that dy/dx = -1/(1+x^{2}).